Bin search
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@ -2,13 +2,16 @@
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#define S0034_FIND_FIRST_AND_LAST_POSITION_OF_ELEMENT_IN_SORTED_ARRAY_HPP
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#define S0034_FIND_FIRST_AND_LAST_POSITION_OF_ELEMENT_IN_SORTED_ARRAY_HPP
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#include <vector>
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#include <vector>
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#include <cmath>
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using namespace std;
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using namespace std;
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class Solution {
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class Solution1 {
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public:
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public:
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vector<int> searchRange(vector<int>& nums, int target);
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vector<int> searchRange(vector<int>& nums, int target);
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};
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};
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class Solution2 {
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public:
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vector<int> searchRange(vector<int>& nums, int target);
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};
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#endif
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#endif
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@ -2,11 +2,15 @@
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#define S0035_SEARCH_INSERT_POSITION_HPP
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#define S0035_SEARCH_INSERT_POSITION_HPP
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#include <vector>
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#include <vector>
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#include <cmath>
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using namespace std;
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using namespace std;
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class Solution {
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class Solution1 {
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public:
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int searchInsert(vector<int>& nums, int target);
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};
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class Solution2 {
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public:
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public:
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int searchInsert(vector<int>& nums, int target);
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int searchInsert(vector<int>& nums, int target);
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};
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};
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18
include/s0704_binary_search.hpp
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18
include/s0704_binary_search.hpp
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#ifndef S0704_BINARY_SEARCH_HPP
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#define S0704_BINARY_SEARCH_HPP
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#include <vector>
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using namespace std;
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class Solution1 {
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public:
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int binSearch(vector<int>& nums, int target);
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};
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class Solution2 {
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public:
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int binSearch(vector<int>& nums, int target);
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};
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#endif
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@ -1,9 +1,9 @@
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# 思路
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# 数组
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- [深度优先遍历](./dfs.md)
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- [二分查找](./bin_search.md)
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- [广度优先遍历](./bfs.md)
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## 经典代码
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# 经典代码
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- [合并两个有序链表](./merge_two_sorted_linked_lists.md)
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- [合并两个有序链表](./merge_two_sorted_linked_lists.md)
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- [二分查找](./bin_search.md)
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- [深度优先遍历](./dfs.md)
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- [广度优先遍历](./bfs.md)
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@ -1,57 +1,64 @@
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# 二分查找
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# 二分查找
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非递归:
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适用于数组有序的情况下查找目标值。
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## 写法一
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针对左闭右闭区间(即 `[left, right]`):
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```cpp
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```cpp
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/**
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class Solution {
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* 二分查找普通实现。
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public:
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* @param srcArray 有序数组
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int search(vector<int>& nums, int target) {
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* @param key 查找元素
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int left = 0;
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* @return 不存在返回-1
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int right = nums.size() - 1; // 定义target在左闭右闭的区间里,[left, right]
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*/
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while (left <=
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public
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right) { // 当left==right,区间[left, right]依然有效,所以用 <=
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static int binSearch(int srcArray[], int key) {
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int middle =
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int mid;
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left + ((right - left) / 2); // 防止溢出 等同于(left + right)/2
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int start = 0;
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if (nums[middle] > target) {
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int end = srcArray.length - 1;
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right = middle - 1; // target 在左区间,所以[left, middle - 1]
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while (start <= end) {
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} else if (nums[middle] < target) {
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mid = (end - start) / 2 + start;
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left = middle + 1; // target 在右区间,所以[middle + 1, right]
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if (key < srcArray[mid]) {
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} else { // nums[middle] == target
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end = mid - 1;
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return middle; // 数组中找到目标值,直接返回下标
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} else if (key > srcArray[mid]) {
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}
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start = mid + 1;
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} else {
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return mid;
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}
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}
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// 未找到目标值
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return -1;
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}
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}
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return -1;
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};
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}
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```
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```
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递归:
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## 写法二
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针对左闭右开(即 `[left, right)`):
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```cpp
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```cpp
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/**
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class Solution {
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* 二分查找递归实现。
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public:
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* @param srcArray 有序数组
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int search(vector<int>& nums, int target) {
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* @param start 数组低地址下标
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int left = 0;
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* @param end 数组高地址下标
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int right = nums.size(); // 定义target在左闭右开的区间里,即:[left, right)
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* @param key 查找元素
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while (left < right) { // 因为left == right的时候,在[left,
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* @return 查找元素不存在返回-1
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// right)是无效的空间,所以使用 <
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*/
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int middle = left + ((right - left) >> 1);
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public
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if (nums[middle] > target) {
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static int binSearch(int srcArray[], int start, int end, int key) {
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right = middle; // target 在左区间,在[left, middle)中
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int mid = (end - start) / 2 + start;
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} else if (nums[middle] < target) {
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if (srcArray[mid] == key) {
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left = middle + 1; // target 在右区间,在[middle + 1, right)中
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return mid;
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} else { // nums[middle] == target
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}
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return middle; // 数组中找到目标值,直接返回下标
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if (start >= end) {
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}
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}
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// 未找到目标值
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return -1;
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return -1;
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} else if (key > srcArray[mid]) {
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return binSearch(srcArray, mid + 1, end, key);
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} else if (key < srcArray[mid]) {
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return binSearch(srcArray, start, mid - 1, key);
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}
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}
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return -1;
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};
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}
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```
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```
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## 相关题目
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- [704. 二分查找](https://programmercarl.com/0704.%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE.html#_704-%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE)
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- [35. 搜索插入位置](https://programmercarl.com/0035.%E6%90%9C%E7%B4%A2%E6%8F%92%E5%85%A5%E4%BD%8D%E7%BD%AE.html#_35-%E6%90%9C%E7%B4%A2%E6%8F%92%E5%85%A5%E4%BD%8D%E7%BD%AE)
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- [34. 在排序数组中查找元素的第一个和最后一个位置](https://programmercarl.com/0034.%E5%9C%A8%E6%8E%92%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9F%A5%E6%89%BE%E5%85%83%E7%B4%A0%E7%9A%84%E7%AC%AC%E4%B8%80%E4%B8%AA%E5%92%8C%E6%9C%80%E5%90%8E%E4%B8%80%E4%B8%AA%E4%BD%8D%E7%BD%AE.html#_34-%E5%9C%A8%E6%8E%92%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9F%A5%E6%89%BE%E5%85%83%E7%B4%A0%E7%9A%84%E7%AC%AC%E4%B8%80%E4%B8%AA%E5%92%8C%E6%9C%80%E5%90%8E%E4%B8%80%E4%B8%AA%E4%BD%8D%E7%BD%AE)
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#include "s0034_find_first_and_last_position_of_element_in_sorted_array.hpp"
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#include "s0034_find_first_and_last_position_of_element_in_sorted_array.hpp"
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vector<int> Solution::searchRange(vector<int>& nums, int target) {
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// 闭区间写法
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int size = nums.size();
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vector<int> Solution1::searchRange(vector<int>& nums, int target) {
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if (size <= 0) {
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int len = nums.size();
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return vector<int> {-1, -1};
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int l{0};
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} else if (size == 1) {
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int r = len - 1;
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if (nums[0] == target) {
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while (l <= r) {
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return vector<int> {0, 0};
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int m = (l + r) >> 1;
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if (target < nums[m]) {
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r = m - 1;
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} else if (nums[m] < target) {
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l = m + 1;
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} else {
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} else {
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return vector<int> {-1, -1};
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l = r = m;
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}
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while (l >= 0) {
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}
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if (nums[l] == target) {
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int left{0};
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--l;
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int right = size - 1;
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} else {
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int mid;
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break;
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while (left <= right) {
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mid = static_cast<int>(floor((left + right) / 2));
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if (nums[mid] == target) {
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int resultLeft, resultRight;
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for (int i = mid; i < size; ++i) {
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if (nums[i] == target) {
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resultRight = i;
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}
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}
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}
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}
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for (int i = mid; i >= 0; --i) {
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while (r <= len - 1) {
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if (nums[i] == target) {
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if (nums[r] == target) {
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resultLeft = i;
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++r;
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} else {
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break;
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}
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}
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}
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}
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return vector<int> {resultLeft, resultRight};
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return vector<int> {l + 1, r - 1};
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} else if (nums[mid] > target) {
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}
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right = mid - 1;
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}
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} else {
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return vector<int> {-1, -1};
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left = mid + 1;
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}
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// 开区间写法
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vector<int> Solution2::searchRange(vector<int>& nums, int target) {
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int len = nums.size();
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int l{0};
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int r = len;
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while (l < r) {
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int m = (l + r) >> 1;
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if (target < nums[m]) {
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r = m;
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} else if (nums[m] < target) {
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l = m + 1;
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} else {
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l = r = m;
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while (l >= 0) {
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if (nums[l] == target) {
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--l;
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} else {
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break;
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}
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}
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while (r <= len - 1) {
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if (nums[r] == target) {
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++r;
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} else {
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break;
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}
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}
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return vector<int> {l + 1, r - 1};
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}
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}
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}
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}
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return vector<int> {-1, -1};
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return vector<int> {-1, -1};
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#include "s0035_search_insert_position.hpp"
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#include "s0035_search_insert_position.hpp"
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int Solution::searchInsert(vector<int>& nums, int target) {
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// 闭区间写法
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int size = nums.size();
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int Solution1::searchInsert(vector<int>& nums, int target) {
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if (size == 0) {
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int len = nums.size();
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return 0;
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int l{0};
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}
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int r = len - 1;
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int left{0};
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while (l <= r) {
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int right = size - 1;
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int m = (l + r) >> 1;
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int mid;
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if (target < nums[m]) {
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while (left <= right) {
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r = m - 1;
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mid = static_cast<int>(floor((left + right) / 2));
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} else if (nums[m] < target) {
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if (nums[mid] == target) {
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l = m + 1;
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return mid;
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} else if (nums[mid] > target) {
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right = mid - 1;
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} else {
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} else {
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left = mid + 1;
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return m;
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}
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}
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}
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}
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if (right < 0) {
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return r + 1;
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return 0;
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}
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} else if (left > size - 1) {
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return size;
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// 开区间写法
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}
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int Solution2::searchInsert(vector<int>& nums, int target) {
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if (target < nums[right]) {
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int len = nums.size();
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return right;
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int l{0};
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} else if (target > nums[left]) {
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int r = len;
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return left + 1;
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while (l < r) {
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} else {
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int m = (l + r) >> 1;
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return left;
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if (target < nums[m]) {
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}
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r = m;
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} else if (nums[m] < target) {
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l = m + 1;
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} else {
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return m;
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}
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}
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return r;
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}
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}
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37
src/s0704_binary_search.cpp
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37
src/s0704_binary_search.cpp
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#include "s0704_binary_search.hpp"
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// 闭区间写法
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int Solution1::binSearch(vector<int>& nums, int target) {
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int len = nums.size();
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int l{0};
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int r = len - 1;
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while (l <= r) {
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int m = (l + r) >> 1;
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if (target < nums[m]) {
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r = m - 1;
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} else if (nums[m] < target) {
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l = m + 1;
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} else {
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return m;
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}
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}
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return -1;
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}
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// 开区间写法
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int Solution2::binSearch(vector<int>& nums, int target) {
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int len = nums.size();
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int l{0};
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int r = len;
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while (l < r) {
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int m = (l + r) >> 1;
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if (target < nums[m]) {
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r = m;
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} else if (nums[m] < target) {
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l = m + 1;
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} else {
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return m;
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}
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}
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return -1;
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}
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@ -6,22 +6,28 @@ TEST(Problem34, Case1) {
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vector<int> nums{5, 7, 7, 8, 8, 10};
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vector<int> nums{5, 7, 7, 8, 8, 10};
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int target{8};
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int target{8};
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vector<int> o{3, 4};
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vector<int> o{3, 4};
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Solution solution;
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Solution1 solution1;
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EXPECT_EQ(solution.searchRange(nums, target), o);
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Solution2 solution2;
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EXPECT_EQ(solution1.searchRange(nums, target), o);
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EXPECT_EQ(solution2.searchRange(nums, target), o);
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}
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}
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TEST(Problem34, Case2) {
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TEST(Problem34, Case2) {
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vector<int> nums{5, 7, 7, 8, 8, 10};
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vector<int> nums{5, 7, 7, 8, 8, 10};
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int target{6};
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int target{6};
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vector<int> o{-1, -1};
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vector<int> o{-1, -1};
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Solution solution;
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Solution1 solution1;
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EXPECT_EQ(solution.searchRange(nums, target), o);
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Solution2 solution2;
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EXPECT_EQ(solution1.searchRange(nums, target), o);
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EXPECT_EQ(solution2.searchRange(nums, target), o);
|
||||||
}
|
}
|
||||||
|
|
||||||
TEST(Problem34, Case3) {
|
TEST(Problem34, Case3) {
|
||||||
vector<int> nums{};
|
vector<int> nums{};
|
||||||
int target{0};
|
int target{0};
|
||||||
vector<int> o{-1, -1};
|
vector<int> o{-1, -1};
|
||||||
Solution solution;
|
Solution1 solution1;
|
||||||
EXPECT_EQ(solution.searchRange(nums, target), o);
|
Solution2 solution2;
|
||||||
|
EXPECT_EQ(solution1.searchRange(nums, target), o);
|
||||||
|
EXPECT_EQ(solution2.searchRange(nums, target), o);
|
||||||
}
|
}
|
||||||
|
@ -6,22 +6,28 @@ TEST(Problem35, Case1) {
|
|||||||
vector<int> nums{1, 3, 5, 6};
|
vector<int> nums{1, 3, 5, 6};
|
||||||
int target{5};
|
int target{5};
|
||||||
int o{2};
|
int o{2};
|
||||||
Solution solution;
|
Solution1 solution1;
|
||||||
EXPECT_EQ(solution.searchInsert(nums, target), o);
|
Solution2 solution2;
|
||||||
|
EXPECT_EQ(solution1.searchInsert(nums, target), o);
|
||||||
|
EXPECT_EQ(solution2.searchInsert(nums, target), o);
|
||||||
}
|
}
|
||||||
|
|
||||||
TEST(Problem35, Case2) {
|
TEST(Problem35, Case2) {
|
||||||
vector<int> nums{1, 3, 5, 6};
|
vector<int> nums{1, 3, 5, 6};
|
||||||
int target{2};
|
int target{2};
|
||||||
int o{1};
|
int o{1};
|
||||||
Solution solution;
|
Solution1 solution1;
|
||||||
EXPECT_EQ(solution.searchInsert(nums, target), o);
|
Solution2 solution2;
|
||||||
|
EXPECT_EQ(solution1.searchInsert(nums, target), o);
|
||||||
|
EXPECT_EQ(solution2.searchInsert(nums, target), o);
|
||||||
}
|
}
|
||||||
|
|
||||||
TEST(Problem35, Case3) {
|
TEST(Problem35, Case3) {
|
||||||
vector<int> nums{1, 3, 5, 6};
|
vector<int> nums{1, 3, 5, 6};
|
||||||
int target{7};
|
int target{7};
|
||||||
int o{4};
|
int o{4};
|
||||||
Solution solution;
|
Solution1 solution1;
|
||||||
EXPECT_EQ(solution.searchInsert(nums, target), o);
|
Solution2 solution2;
|
||||||
|
EXPECT_EQ(solution1.searchInsert(nums, target), o);
|
||||||
|
EXPECT_EQ(solution2.searchInsert(nums, target), o);
|
||||||
}
|
}
|
||||||
|
23
tests/s0704_binary_search.cpp
Normal file
23
tests/s0704_binary_search.cpp
Normal file
@ -0,0 +1,23 @@
|
|||||||
|
#include "s0704_binary_search.hpp"
|
||||||
|
|
||||||
|
#include <gtest/gtest.h>
|
||||||
|
|
||||||
|
TEST(Problem704, Case1) {
|
||||||
|
vector<int> nums{-1, 0, 3, 5, 9, 12};
|
||||||
|
int target{9};
|
||||||
|
int o{4};
|
||||||
|
Solution1 solution1;
|
||||||
|
Solution2 solution2;
|
||||||
|
EXPECT_EQ(solution1.binSearch(nums, target), o);
|
||||||
|
EXPECT_EQ(solution2.binSearch(nums, target), o);
|
||||||
|
}
|
||||||
|
|
||||||
|
TEST(Problem704, Case2) {
|
||||||
|
vector<int> nums{-1, 0, 3, 5, 9, 12};
|
||||||
|
int target{2};
|
||||||
|
int o{-1};
|
||||||
|
Solution1 solution1;
|
||||||
|
Solution2 solution2;
|
||||||
|
EXPECT_EQ(solution1.binSearch(nums, target), o);
|
||||||
|
EXPECT_EQ(solution2.binSearch(nums, target), o);
|
||||||
|
}
|
Loading…
Reference in New Issue
Block a user