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.
从两边向中间考虑时,乘水的面积是由(两端较小的高度)×(两个板之间的宽度)决定的。记录最开始的乘水面积为ans1,然后L向右运动,R向左运动,截止条件是L >= R,并且记录乘水的面积ans,取最大值
以L向左运动为例,当宽度减小时,如果面积变大,必然高度要增加,因此L只需取比前一个L大的值即可,初始L的高度为L1。R向右运动同理
public class Solution { public int maxArea(int[] height) { if(height.length<2) return 0; int re=0; int j=height.length-1; int i=0; while(j>i){ if(height[i]>=height[j]){ int area = height[j]*(j-i); if(area>re) re=area; j--; continue; }else{ int area=height[i]*(j-i); if(area>re) re=area; i++; continue; } } return re; } }