s0011
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Sainnhe Park 2022-11-03 15:22:56 +08:00
parent 914c04c9c5
commit 090003819d
3 changed files with 54 additions and 0 deletions

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#ifndef S0011_CONTAINER_WITH_MOST_WATER
#define S0011_CONTAINER_WITH_MOST_WATER
#include <vector>
using namespace std;
class Solution {
public:
int maxArea(vector<int>& height);
};
#endif

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#include "s0011_container_with_most_water.hpp"
// 双指针:
// 一开始左指针指向数组开头,右指针指向数组结尾
// 两个指针向中间移动
// 每次移动的时候移动较小的那个指针:
// 由于储存的水量 = 两指针之间的距离 * 较小的指针的值
// 因此如果移动较大的指针,那么储存的水量一定不会超过之前的
// 也就是说只有当移动小的指针时才有可能让储存的水量超过之前的
// 因此每次迭代我们移动较小的指针,然后统计出最大的储存水量
int Solution::maxArea(vector<int>& height) {
int max{0};
for (int l{0}, r{static_cast<int>(height.size() - 1)}; r != l;) {
if (height.at(l) > height.at(r)) {
max = (r - l) * height.at(r) > max ? (r - l) * height.at(r) : max;
r--;
} else {
max = (r - l) * height.at(l) > max ? (r - l) * height.at(l) : max;
l++;
}
}
return max;
}

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#include "s0011_container_with_most_water.hpp"
#include <gtest/gtest.h>
TEST(Problem11, Case1) {
vector<int> height{1, 8, 6, 2, 5, 4, 8, 3, 7};
int o{49};
Solution solution;
EXPECT_EQ(solution.maxArea(height), o);
}
TEST(Problem11, Case2) {
vector<int> height{1, 1};
int o{1};
Solution solution;
EXPECT_EQ(solution.maxArea(height), o);
}