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!
class Solution { public: int trap(int A[], int n) { if(NULL == A || 0 == n) return 0; vector<int> left(n); int maxHeight = 0; for(int i = 0; i < n; ++i) { left[i] = maxHeight; maxHeight = max(maxHeight, A[i]); } maxHeight = 0; vector<int> right(n); for(int i = n - 1; i >= 0; --i) { right[i] = maxHeight; maxHeight = max(maxHeight, A[i]); } int sum = 0; int height = 0; for(int i = 0; i < n; ++i) { height = min(left[i], right[i]) - A[i]; sum += height > 0? height: 0; } return sum; } };