Container With Most Water问题及解法

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问题描述：

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

注意：

You may not slant the container and n is at least 2.

问题分析：

假设从头到尾最大宽度的容器A包含最多的水，若要想找到比它装水还多的容器B，则该容器B的高度肯定要比A的高度要高。

详细过程可见代码：

class Solution {
public:int maxArea(vector<int>& height) {int water = 0;int i = 0, j = height.size() - 1;while(i < j){int h = min(height[i], height[j]);water = max(water, (j - i) * h);while(height[i] <= h && i < j) i++;while(height[j] <= h && i < j) j--;}return water;}
};

代码还是比较容易理解的~~

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