Trapping Rain Water

最新推荐文章于 2025-09-10 21:59:39 发布

转载最新推荐文章于 2025-09-10 21:59:39 发布 · 40 阅读

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原文链接：http://www.cnblogs.com/amazingzoe/p/4783934.html

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#runtime

本文介绍了一个算法，用于计算给定高度数组表示的地形中，在雨后能够储存的雨水量。通过找到每个元素左侧和右侧的最大高度，算法计算了能够储存的雨水量。实现效率高，运行速度快。

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!

Analyse: The water can be trapped equals to the min(left[i], right[i]) - height[i], whereas left[i] and right[i] is the largest number of the left and right part of the number.

Runtime: 8ms.

 1 class Solution {
 2 public:
 3     int trap(vector<int>& height) {
 4         int n = height.size();
 5         if(n <= 1) return 0;
 6         
 7         int *left = new int [n];
 8         int max = height[0];
 9         for(int i = 1; i < n - 1; i++){//find the largest of the left of a element
10             left[i] = max;
11             if(height[i] > max) 
12                 max = height[i];
13         }
14         max = height[n - 1];
15         int result = 0;
16         for(int j = n - 2; j >= 0; j--){
17             int diff = min(max, left[j]) - height[j]; //the water trapped at index i
18             if(diff > 0) 
19                 result += diff;
20             if(height[j] > max) 
21                 max = height[j];
22         }
23         return result;
24     }
25 };