题目描述
Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container and n is at least 2.
解题思想
关键的思路在于:对于任意的容器,它的容积取决于最短的那个板子。
采用双指针的思想,从首尾往中间扫描数组,扫描过程中,只移动短板端的指针,因为容器的容积取决于短板,只有在改变短板的时候才有可能产生新的最大值。
解题代码
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
int maxArea(vector<int>& height) {
int left = 0;
int right = height.size() - 1;
int maxWater = 0;
while(left < right){
int temp = min(height[left], height[right]) * (right - left);
maxWater = max(maxWater, temp);
if(height[left] < height[right])
left++;
else
right--;
}
return maxWater;
}
};
int main()
{
Solution solution;
vector<int> height = {7, 5, 6, 9};
cout << solution.maxArea(height) << endl;
return 0;
}