LeetCode--Container With Most Water-优快云博客

本文探讨了如何通过计算垂直线形成的容器中所能容纳的最大水量。利用双指针技巧优化了搜索过程，确保了在不倾斜容器的前提下找到最优解。

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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.

解题思路：影响蓄水量的因素包括容器的宽度以及容器的最短的挡板的高度，因此我们可以先将容器的宽度设为最大宽度，如此即可得到此时的最大值，为了让容器的体积增大，只有将最短的边变高才可能在容器宽度减小的情况下增大容器的容量。因此即可得到如下的代码

class Solution 
{
public:
    int maxArea(vector<int> &height) 
    {
        int n = height.size();
        if(n<=1)
            return 0;
        vector<int> h = height;
        int start=0;
        int end = n-1;
        int res = 0;
        while(start<end)
        {
            int temp = end - start;
            if(h[start] > h[end])
            {
                temp = temp*h[end];
                end--;
            }
            else 
            {
                temp = temp*h[start];
                start++;
            }
            if(temp > res)
                res = temp;
        }
        return res;
    }
};