LeetCode 11. Container With Most Water

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 and n is at least 2.
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题目

给定 n 个非负整数 a1, a2, …, an, 每个数代表了坐标中的一个点 (i, ai)。画 n 条垂直线,使得 i 垂直线的两个端点分别为(i, ai)和(i, 0)。找到两条线,使得其与 x 轴共同构成一个容器,以容纳最多水。
注意事项
容器不可倾斜。
样例
给出[1,3,2], 最大的储水面积是2

分析

用两根指针一头一尾,分别计算面积,然后选取高度小的那边向中间移动,因为这样才可能存在更大的面积,更新最大的面积,最后就是结果。

《LeetCode 11. Container With Most Water》 Paste_Image.png

代码

public class Solution {
    /**
     * @param heights: an array of integers
     * @return: an integer
     */
    int computeArea(int left, int right,  int[] heights) {
        return (right-left)*Math.min(heights[left], heights[right]);
    }
    
    public int maxArea(int[] heights) {
        //两根指针,一头一尾计算,短的那根指针向中间移动,因为只有这种情况才存在更大的情况
        int left = 0, right = heights.length-1;
        int res = 0;
        
        while(left < right) {
            res = Math.max(res, computeArea(left,right,heights));
            if(heights[left] < heights[right]) {
                left++;
                //如果小于则不必要计算,直接跳过
                while(heights[left] <= heights[left-1] && left<right)
                    left++;
            }
            else {
                right--;
                while(heights[right] <= heights[right+1] && left <right)
                    right--;
            }
        }
        return res; 
    }
}
    原文作者:六尺帐篷
    原文地址: https://www.jianshu.com/p/3056972e1c92
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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