Leetcode——42. Trapping Rain Water(C++)

Leetcode——42. Trapping Rain Water(C++)
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.

Example 1:
Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) 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.

Example 2:
Input: height = [4,2,0,3,2,5]
Output: 9

Constraints:
n == height.length
1 <= n <= 2 * 104
0 <= height[i] <= 105

解题思路：记录每点左边最高值和右边最高值，二者较小者减去此点高度即是此点可蓄水值，此点高于左右最高值则此点不蓄水；

代码；

class Solution {
public:
    int trap(vector& height) {
        int siz=height.size();
        vectorleftmax(siz,0),rightmax(siz,0);
        int lem=0,rma=0;
        for(int i=0;i
            leftmax[i]=lem;
            if(height[i]>lem)lem=height[i];
        }
        for(int j=siz-1;j>=0;j--)
        {
            rightmax[j]=rma;
            if(height[j]>rma)rma=height[j];
        }
        int sum=0;
        for(int i=1;i
            if(height[i] 
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