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.

看了nandawys的评论，找到了O(n)方法，思路是从两头到中间扫描，设i,j分别指向height数组的首尾。
那么当前的area是min(height[i],height[j]) * (j-i)。
当height[i] < height[j]的时候，我们把i往后移，否则把j往前移，直到两者相遇。

public int maxArea(int[] height) {
        int start = 0;
        int end = height.length - 1;
        int max = 0;
        while(start < end){
            int container = Math.min(height[start], height[end]) * (end - start);
            if(max < container){
                max = container;
            }
            
            if(height[start] > height[end]){
                end --;
            } else {
                start ++;
            }
        }
        
        return max;
    }

Container With Most Water

原文：http://www.cnblogs.com/RazerLu/p/3552511.html