42-Trapping Rain Water

最新推荐文章于 2021-03-29 02:37:52 发布

转载最新推荐文章于 2021-03-29 02:37:52 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/hwu2014/p/4481530.html

本文介绍了一种计算雨后地面积水总量的高效算法。通过分析地形高度数据，利用两边逼近的方法并记录第二高点来计算能容纳多少雨水。以数组[0,1,0,2,1,0,1,3,2,1,2,1]为例，展示了如何实现该算法并通过实例返回6单位的积水结果。

【题目】

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

【analyze】

1.两边逼近

2.记录第二高点

【算法】

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