盛水最大量---贪心算法

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.

public class Solution {    public int maxArea(int[] height) {        int left = 0, right = height.length - 1, maxArea = 0;        while(left < right){            // 每次更新最大面积（盛水量）            maxArea = Math.max(maxArea,(right-left)* Math.min(height[left], height[right]));            // 如果左边较低，则将左边向中间移一点            if(height[left] < height[right]){                left++;            // 如果右边较低，则将右边向中间移一点            } else {                right--;            }        }        return maxArea;    }}