Bin search

include/s0034_find_first_and_last_position_of_element_in_sorted_array.hpp

@@ -2,13 +2,16 @@
 #define S0034_FIND_FIRST_AND_LAST_POSITION_OF_ELEMENT_IN_SORTED_ARRAY_HPP
 
 #include <vector>
-#include <cmath>
 
 using namespace std;
 
-class Solution {
+class Solution1 {
  public:
   vector<int> searchRange(vector<int>& nums, int target);
 };
 
+class Solution2 {
+ public:
+  vector<int> searchRange(vector<int>& nums, int target);
+};
 #endif

include/s0035_search_insert_position.hpp

@@ -2,11 +2,15 @@
 #define S0035_SEARCH_INSERT_POSITION_HPP
 
 #include <vector>
-#include <cmath>
 
 using namespace std;
 
-class Solution {
+class Solution1 {
+ public:
+  int searchInsert(vector<int>& nums, int target);
+};
+
+class Solution2 {
  public:
   int searchInsert(vector<int>& nums, int target);
 };

include/s0704_binary_search.hpp

@@ -0,0 +1,18 @@
+#ifndef S0704_BINARY_SEARCH_HPP
+#define S0704_BINARY_SEARCH_HPP
+
+#include <vector>
+
+using namespace std;
+
+class Solution1 {
+ public:
+  int binSearch(vector<int>& nums, int target);
+};
+
+class Solution2 {
+ public:
+  int binSearch(vector<int>& nums, int target);
+};
+
+#endif

notes/src/SUMMARY.md

@@ -1,9 +1,9 @@
-# 思路
+# 数组
 
-- [深度优先遍历](./dfs.md)
+- [二分查找](./bin_search.md)
-- [广度优先遍历](./bfs.md)
 
-## 经典代码
+# 经典代码
 
 - [合并两个有序链表](./merge_two_sorted_linked_lists.md)
-- [二分查找](./bin_search.md)
+- [深度优先遍历](./dfs.md)
+- [广度优先遍历](./bfs.md)

notes/src/bin_search.md

@@ -1,57 +1,64 @@
 # 二分查找
 
-非递归：
+适用于数组有序的情况下查找目标值。
+
+## 写法一
+
+针对左闭右闭区间（即 `[left, right]`）：
 
 ```cpp
-/**
+class Solution {
- * 二分查找普通实现。
+ public:
- * @param srcArray 有序数组
+  int search(vector<int>& nums, int target) {
- * @param key 查找元素
+    int left = 0;
- * @return  不存在返回-1
+    int right = nums.size() - 1;  // 定义target在左闭右闭的区间里，[left, right]
- */
+    while (left <=
-public
+           right) {  // 当left==right，区间[left, right]依然有效，所以用 <=
-static int binSearch(int srcArray[], int key) {
+      int middle =
-  int mid;
+          left + ((right - left) / 2);  // 防止溢出 等同于(left + right)/2
-  int start = 0;
+      if (nums[middle] > target) {
-  int end = srcArray.length - 1;
+        right = middle - 1;  // target 在左区间，所以[left, middle - 1]
-  while (start <= end) {
+      } else if (nums[middle] < target) {
-    mid = (end - start) / 2 + start;
+        left = middle + 1;  // target 在右区间，所以[middle + 1, right]
-    if (key < srcArray[mid]) {
+      } else {              // nums[middle] == target
-      end = mid - 1;
+        return middle;      // 数组中找到目标值，直接返回下标
-    } else if (key > srcArray[mid]) {
-      start = mid + 1;
-    } else {
-      return mid;
       }
     }
+    // 未找到目标值
     return -1;
   }
+};
 ```
 
-递归：
+## 写法二
+
+针对左闭右开（即 `[left, right)`）：
 
 ```cpp
-/**
+class Solution {
- * 二分查找递归实现。
+ public:
- * @param srcArray  有序数组
+  int search(vector<int>& nums, int target) {
- * @param start 数组低地址下标
+    int left = 0;
- * @param end   数组高地址下标
+    int right = nums.size();  // 定义target在左闭右开的区间里，即：[left, right)
- * @param key  查找元素
+    while (left < right) {  // 因为left == right的时候，在[left,
- * @return 查找元素不存在返回-1
+                            // right)是无效的空间，所以使用 <
- */
+      int middle = left + ((right - left) >> 1);
-public
+      if (nums[middle] > target) {
-static int binSearch(int srcArray[], int start, int end, int key) {
+        right = middle;  // target 在左区间，在[left, middle)中
-  int mid = (end - start) / 2 + start;
+      } else if (nums[middle] < target) {
-  if (srcArray[mid] == key) {
+        left = middle + 1;  // target 在右区间，在[middle + 1, right)中
-    return mid;
+      } else {              // nums[middle] == target
+        return middle;      // 数组中找到目标值，直接返回下标
       }
-  if (start >= end) {
-    return -1;
-  } else if (key > srcArray[mid]) {
-    return binSearch(srcArray, mid + 1, end, key);
-  } else if (key < srcArray[mid]) {
-    return binSearch(srcArray, start, mid - 1, key);
     }
+    // 未找到目标值
     return -1;
   }
+};
 ```
+
+## 相关题目
+
+- [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)
+- [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)
+- [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)

src/s0034_find_first_and_last_position_of_element_in_sorted_array.cpp

@@ -1,38 +1,66 @@
 #include "s0034_find_first_and_last_position_of_element_in_sorted_array.hpp"
 
-vector<int> Solution::searchRange(vector<int>& nums, int target) {
+// 闭区间写法
-  int size = nums.size();
+vector<int> Solution1::searchRange(vector<int>& nums, int target) {
-  if (size <= 0) {
+  int len  = nums.size();
-    return vector<int> {-1, -1};
+  int l{0};
-  } else if (size == 1) {
+  int r = len - 1;
-    if (nums[0] == target) {
+  while (l <= r) {
-      return vector<int> {0, 0};
+    int m = (l + r) >> 1;
+    if (target < nums[m]) {
+      r = m - 1;
+    } else if (nums[m] < target) {
+      l = m + 1;
     } else {
-      return vector<int> {-1, -1};
+      l = r = m;
-    }
+      while (l >= 0) {
-  }
+        if (nums[l] == target) {
-  int left{0};
+          --l;
-  int right = size - 1;
-  int mid;
-  while (left <= right) {
-    mid = static_cast<int>(floor((left + right) / 2));
-    if (nums[mid] == target) {
-      int resultLeft, resultRight;
-      for (int i = mid; i < size; ++i) {
-        if (nums[i] == target) {
-          resultRight = i;
-        }
-      }
-      for (int i = mid; i >= 0; --i) {
-        if (nums[i] == target) {
-          resultLeft = i;
-        }
-      }
-      return vector<int> {resultLeft, resultRight};
-    } else if (nums[mid] > target) {
-      right = mid - 1;
         } else {
-      left = mid + 1;
+          break;
+        }
+      }
+      while (r <= len - 1) {
+        if (nums[r] == target) {
+          ++r;
+        } else {
+          break;
+        }
+      }
+      return vector<int> {l + 1, r - 1};
+    }
+  }
+  return vector<int> {-1, -1};
+}
+
+// 开区间写法
+vector<int> Solution2::searchRange(vector<int>& nums, int target) {
+  int len  = nums.size();
+  int l{0};
+  int r = len;
+  while (l < r) {
+    int m = (l + r) >> 1;
+    if (target < nums[m]) {
+      r = m;
+    } else if (nums[m] < target) {
+      l = m + 1;
+    } else {
+      l = r = m;
+      while (l >= 0) {
+        if (nums[l] == target) {
+          --l;
+        } else {
+          break;
+        }
+      }
+      while (r <= len - 1) {
+        if (nums[r] == target) {
+          ++r;
+        } else {
+          break;
+        }
+      }
+      return vector<int> {l + 1, r - 1};
     }
   }
   return vector<int> {-1, -1};

src/s0035_search_insert_position.cpp

@@ -1,33 +1,37 @@
 #include "s0035_search_insert_position.hpp"
 
-int Solution::searchInsert(vector<int>& nums, int target) {
+// 闭区间写法
-  int size = nums.size();
+int Solution1::searchInsert(vector<int>& nums, int target) {
-  if (size == 0) {
+  int len = nums.size();
-    return 0;
+  int l{0};
-  }
+  int r = len - 1;
-  int left{0};
+  while (l <= r) {
-  int right = size - 1;
+    int m = (l + r) >> 1;
-  int mid;
+    if (target < nums[m]) {
-  while (left <= right) {
+      r = m - 1;
-    mid = static_cast<int>(floor((left + right) / 2));
+    } else if (nums[m] < target) {
-    if (nums[mid] == target) {
+      l = m + 1;
-      return mid;
-    } else if (nums[mid] > target) {
-      right = mid - 1;
     } else {
-      left = mid + 1;
+      return m;
     }
   }
-  if (right < 0) {
+  return r + 1;
-    return 0;
-  } else if (left > size - 1) {
-    return size;
 }
-  if (target < nums[right]) {
+
-    return right;
+// 开区间写法
-  } else if (target > nums[left]) {
+int Solution2::searchInsert(vector<int>& nums, int target) {
-    return left + 1;
+  int len = nums.size();
+  int l{0};
+  int r = len;
+  while (l < r) {
+    int m = (l + r) >> 1;
+    if (target < nums[m]) {
+      r = m;
+    } else if (nums[m] < target) {
+      l = m + 1;
     } else {
-    return left;
+      return m;
     }
   }
+  return r;
+}

src/s0704_binary_search.cpp

@@ -0,0 +1,37 @@
+#include "s0704_binary_search.hpp"
+
+// 闭区间写法
+int Solution1::binSearch(vector<int>& nums, int target) {
+  int len = nums.size();
+  int l{0};
+  int r = len - 1;
+  while (l <= r) {
+    int m = (l + r) >> 1;
+    if (target < nums[m]) {
+      r = m - 1;
+    } else if (nums[m] < target) {
+      l = m + 1;
+    } else {
+      return m;
+    }
+  }
+  return -1;
+}
+
+// 开区间写法
+int Solution2::binSearch(vector<int>& nums, int target) {
+  int len = nums.size();
+  int l{0};
+  int r = len;
+  while (l < r) {
+    int m = (l + r) >> 1;
+    if (target < nums[m]) {
+      r = m;
+    } else if (nums[m] < target) {
+      l = m + 1;
+    } else {
+      return m;
+    }
+  }
+  return -1;
+}

tests/s0034_find_first_and_last_position_of_element_in_sorted_array.cpp

@@ -6,22 +6,28 @@ TEST(Problem34, Case1) {
   vector<int> nums{5, 7, 7, 8, 8, 10};
   int target{8};
   vector<int> o{3, 4};
-  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);
 }
 
 TEST(Problem34, Case2) {
   vector<int> nums{5, 7, 7, 8, 8, 10};
   int target{6};
   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);
 }
 
 TEST(Problem34, Case3) {
   vector<int> nums{};
   int target{0};
   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);
 }

tests/s0035_search_insert_position.cpp

@@ -6,22 +6,28 @@ TEST(Problem35, Case1) {
   vector<int> nums{1, 3, 5, 6};
   int target{5};
   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) {
   vector<int> nums{1, 3, 5, 6};
   int target{2};
   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) {
   vector<int> nums{1, 3, 5, 6};
   int target{7};
   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);
 }

tests/s0704_binary_search.cpp

@@ -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);
+}