Container With Most Water

    public int maxArea(int[] height) {
        //http://blog.csdn.net/linhuanmars/article/details/21145429 贪心算法要看看
        if(height==null || height.length<2) return 0;
        int max = 0;
        int l=0, h = height.length-1;
        while(l<h){
            max = Math.max(max,Math.min(height[l], height[h])*(h-l));
            if(height[l]<height[h]){
                l++;
            }else{
                h--;
            }
        }
        return max;
        
    }

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.

 

题解抄自code ganker

“首先一般我们都会想到brute force的方法，思路很简单，就是对每一对pair都计算一次容积，然后去最大的那个，总共有n*(n-1)/2对pair，所以时间复杂度是O(n^2)。 
接下来我们考虑如何优化。思路有点类似于Two Sum中的第二种方法--夹逼。从数组两端走起，每次迭代时判断左pointer和右pointer指向的数字哪个大，如果左pointer小，意味着向左移动右pointer不可能使结果变得更好，因为瓶颈在左pointer，移动右pointer只会变小，所以这时候我们选择左pointer右移。反之，则选择右pointer左移。在这个过程中一直维护最大的那个容积。”