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!
有点类似于最大子矩阵问题：
class Solution {
public:
int trap(int A[], int n) {
if(!n)return 0;
vector<int> L(n);
vector<int> R(n,n);
int result=0;
int maxH=0;
for(int i=0;i<n;i++){
L[i]=maxH;
maxH=max(L[i],A[i]);
}
R[n-1]=0;
maxH=A[n-1];
for(int i=n-2;i>0;i--){
R[i]=maxH;
maxH=max(R[i],A[i]);
int height=min(L[i],R[i]);
if(height>A[i]){
result+=height-A[i];
}
}
return result;
}
};LeetCode OJ:Trapping Rain Water
原文：http://blog.csdn.net/starcuan/article/details/18865617