[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;
}
};