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!
分析: 对于每一点只需要知道其左边和右边的最高山即可
class Solution { public: int trap(int A[], int n) { // Start typing your C/C++ solution below // DO NOT write int main() function if(n <= 2) return 0; vector<int> left(n,0); vector<int> right(n, 0); left[0] = A[0]; for(int i =1; i < n;++i) { left[i] = left[i-1] > A[i] ? left[i-1] : A[i]; } right[n-1] = A[n-1]; for(int i = n-2; i>=0 ;--i){ right[i] = right[i+1] > A[i] ? right[i+1] : A[i]; } int res = 0; for(int i = 0; i< n; i++){ int min = left[i] < right[i]? left[i] : right[i]; res += min - A[i]; } return res; } };