[Leetcode]Trapping Rain Water

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!

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class Solution {public:    /*algorithm two pointer        1)select range[i,j] which statisfy for each i < k < j, h(k) <= min(h(i),h(j))        2)scan array from start to end to compute all the ranges value        time O(n) space O(n)    */    int trap(vector<int>& height) {        int n = height.size();        if(n < 1)return 0;        vector<int>maxH(n,0);        maxH[n-1]=height[n-1];        for(int i = n-2;i >= 0;i--){            maxH[i] = max(maxH[i+1],height[i]);        }        int i = 0,j,sum = 0;        while(i < n){            j = i+1;            if(j < n && maxH[j] > 0){                int h = min(height[i],maxH[j]);                while(j < n && height[j] < h){                    sum += h - min(h,height[j]);                    ++j;                }            }            i = j;        }        return sum;    }};