LEETCODE: Container With Most Water

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.

如何找到这个最大值，我原来想了一个n平方的算法，不过没有通过LEETCODE的测试。所以，必须找到一个时间效率更高的算法。

假设我们用两个指针，一个指向左边left，一个指向右边right。这个时候我们会有一个初始的容量。如果left的高度小于right的高度，那么我们知道right往左移动是不会得到我们想要的，只有left往右移动。（如果我们把right往左移动，容量是right新的位置和left中的小值乘上距离。）反之，right往左移动。

class Solution {public:    int maxArea(vector<int> &height) {        int left = 0;        int right = height.size() - 1;        int max = 0;        while(left < right) {            int length = right - left;            int temp = length * (height[left] < height[right] ? height[left] : height[right]);            max = max > temp ? max : temp;            if(height[left] < height[right])                left ++;            else                right --;        }                return max;    }};