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;
}
}