LeetCode11. Container With Most Water

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题目:

  Given n non-negative integers a1a2, ..., an, where each represents a point at coordinate (iai). n vertical lines are drawn such that the two endpoints of line i is at (iai) 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.

翻译:

  给定n个非负整数a1,a2,...,an,其中每个代表一个点坐标(i,ai)。 n个垂直线段例如线段的两个端点在(i,ai)和(i,0)。 找到两个线段,与x轴形成一个容器,使其包含最多的水。

  备注:不考虑倾斜容器。

思路:容积其实就是面积,我们都知道长方形的面积=长*宽,不妨我们就从长最长的长方形找起,即令left = 0, right = height.size() - 1,但是在找下一个长方形时,长肯定会变短,要弥补这一段损失就必须加宽宽度,所以一下个就换掉两条宽中较小的那一个。

public class Solution {
    public int maxArea(int[] height) {
        int left=0;
        int right=height.length-1;
        int maxArea=0;
        while(left<right){
            maxArea=Math.max(maxArea,Math.min(height[left],height[right])*(right-left));
            if(height[left]>height[right]){
                right--;
            }else{
                left++;
            }
        }
        return maxArea;
    }
}

  

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