22.ContainerWithMostWater（能装最多水的容器）

Level：

??Medium

题目描述：

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 and n is at least 2.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

思路分析：

??从两端往中间搜索，面积的计算方法为两端中较小的作为高，坐标差作为宽，然后求积就是面积，时间复杂度为O(n)。

代码：

public class Solution{
public int maxArea(int []height){
int i=0; //左端
int j=height.length-1;
int maxarea=Math.min(height[i],height[j])*(j-i);
while(i if(height[i]<=height[j])
i++; //为什么要i++而不是j++，因为要尽可能保留高度高的
else
j--;
int area=Math.min(height[i],height[j])*(j-i);
maxarea=Math.max(maxarea,area);
}
return maxarea;
}
}