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. 

这个题目比较简单，常规的思想很容易编写的，所有的情况遍历一遍就行了。但是当边的数量过多的时候，时间过长，毕竟是平方量级的。

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

所以，这道题的关键在于如何减小边的选择。因此，从底最宽的情况开始计算，保留高度较高的边，逐渐向中间缩减。就可以将复杂度降为一次。代码如下：

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