leetcode栈--4、trapping-rain-water（接雨水）

题目描述

 

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], return6.

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!

解题思路：采用dp的思想，定义一个数组，

1）先从左向右遍历一遍，针对每一个点，求出它左侧最大点，记录在数组中，如图记录结果为 001122223333

2）再从右向左遍历，针对每一个点，求出其右侧的最大值（左右侧最小值记录在数组中，短板原理），记录结果为  012222221100(B[11]-B[0])

3）针对每一个点，如果B[i]>A[i]则说明该点可存水(B[i]-A[i])*1

则求出area

 1 class Solution {
 2 public:
 3     int trap(int A[], int n) {
 4         if(A == NULL || n<=0)
 5             return 0;
 6         int area = 0;
 7         int B[n];
 8         int maxVal = 0;
 9         //从左向右遍历，记录每个点左侧的最大值
10         for(int i=0;i)
11         {
12             B[i] = maxVal;
13             maxVal = max(maxVal,A[i]);
14         }
15         maxVal = 0;//重置
16         //从右向左遍历，找每个点右侧的最大值
17         for(int i=n-1;i>=0;i--)
18         {
19             //短板原理，取左右最大值中最小的
20             B[i] = min(maxVal,B[i]);
21             maxVal = max(maxVal,A[i]);
22             if(B[i]>A[i])
23                 area+=(B[i]-A[i]);
24         }
25         return area;
26     }
27 };

 

转载于:https://www.cnblogs.com/qqky/p/6953713.html