Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

https://oj.leetcode.com/problems/container-with-most-water/

思路1：O(n^2)复杂穷举所有的情况，too young too simple.

思路2：:两个指针i,j分别指向首位，选择二者中小的向中间移动，同时更新最大值。这个想法的基础是，如果i的长度小于j，如果移动j，短板在i，不可能找到比当前记录的area更大的值了，只能通过移动i来找到新的可能的更大面积。

public class Solution {    public int maxArea(int[] height) {        int n = height.length;        int maxA = 0;        int curA = 0;        int i = 0, j = n - 1;        while (i < j) {            if (height[i] <= height[j]) {                curA = Math.abs(j - i) * height[i];                if (curA > maxA)                    maxA = curA;                i++;            } else {                curA = Math.abs(j - i) * height[j];                if (curA > maxA)                    maxA = curA;                j--;            }        }        return maxA;    }    public static void main(String[] args) {        System.out.println(new Solution().maxArea(new int[] { 2, 2, 2 }));        System.out.println(new Solution().maxArea(new int[] { 1, 2, 3 }));        System.out.println(new Solution().maxArea(new int[] { 2, 3, 3 }));    }}

参考：

http://blog.csdn.net/wzy_1988/article/details/17248209

http://www.cnblogs.com/zhaolizhen/p/Containerwater.html