### LeetCode

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.

1）两个for循环，遍历所有情况，时间O(n^2). 超时了.

`package area;public class ContainerWithMostWater {    public int maxArea(int[] height) {        int n = height.length;        int maxArea = 0;        for (int i = 0; i < n - 1; ++i) {            for (int j = i + 1; j < n; ++i) {                int area = Math.min(height[i], height[j]) * (j - i);                if (maxArea < area)                    maxArea = area;            }        }                return maxArea;    }        public static void main(String[] args) {        // TODO Auto-generated method stub    }}`

2）用两边夹的方式解决。设Aij为坐标(ai, aj)所圈定的区域面积，其中j > i，则Aij = min(ai, aj)*(j - i)。

如果ai <= aj，则对于k < j，有Aik = min(ai, ak) * (k - i)。由于(k - i) < (j - i)，min(ai, ak) <= min(ai, aj) ，所以Aik < Aij. 这说明，j没必要往左移，这是只能让i往右移。

如果ai > aj，则对于k < i，有Aik = min(ak, aj) * (j - k)。由于(j - k) < (j - i)，min(ak, aj) <= min(ai, aj) ，所以Aik < Aij. 这说明，i没必要往右移，这是只能让j往左移。

`package area;public class ContainerWithMostWater {    public int maxArea(int[] height) {        int n = height.length;        int maxArea = 0;        int i = 0;        int j = n - 1;        while (i < j) {            int area = Math.min(height[i], height[j]) * (j - i);            if (height[i] <= height[j])                 ++i;            else                 --j;                                    if (area > maxArea)                maxArea = area;        }                return maxArea;    }        public static void main(String[] args) {        // TODO Auto-generated method stub    }}`