LeetCode 11. Container With Most Water

暴力O(n^2)，但是可以通过一些“剪枝”使其达到O(n)，但是证明是一个问题。

We starts with the widest container, l = 0 and r = n - 1. Let's say the left one is shorter: h[l] < h[r]. Then, this is already the largest container the left one can form. There's no need to consider it again. Therefore, we just throw it away and start again with l = 1 and r = n -1.

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