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

class Solution {
public:
    int trap(int A[], int n) {
        if(A == NULL || n<=0)
            return 0;
        int area = 0;
        int B[n];
        int maxVal = 0;
        //从左向右遍历，记录每个点左侧的最大值
        for(int i=0;i<n;i++)
        {
            B[i] = maxVal;
            maxVal = max(maxVal,A[i]);
        }
        maxVal = 0;//重置
        //从右向左遍历，找每个点右侧的最大值
        for(int i=n-1;i>=0;i--)
        {
            //短板原理，取左右最大值中最小的
            B[i] = min(maxVal,B[i]);
            maxVal = max(maxVal,A[i]);
            if(B[i]>A[i])
                area+=(B[i]-A[i]);
        }
        return area;
    }
};