南宁网站公司广州疫情最新情况

题目传送门 

1.首先计算出暂时的盛水体积

2.求暂时体积和最大体积max的最大值

3.更新right和left。如果height[left] > height[right] 那么right--否则left++;

class Solution {public int maxArea(int[] height) {int left = 0,right = height.length-1; int ret = 0;while (left < right){/*** 暂时的容量*/int temp = Math.min(height[left],height[right]) * (right-left);/*** 更新容量，找到最大值*/ret = Math.max(temp,ret);/****/if(height[left] > height[right]){right--;}else {left++;}}return ret;}
}