[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!

 1 class Solution {
 2 public:
 3     int trap(int A[], int n) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function
 6         vector<int> left(n);
 7         int maxHeight = 0;
 8         for(int i = 0; i < n; i++)
 9         {
10             left[i] = maxHeight;
11             maxHeight = max(maxHeight, A[i]);
12         }
13         
14         vector<int> right(n);
15         maxHeight = 0;
16         for(int i = n - 1; i >= 0; i--)
17         {
18             right[i] = maxHeight;
19             maxHeight = max(maxHeight, A[i]);
20         }
21         
22         int water = 0;
23         for(int i = 0; i < n; i++)
24         {
25             int height = min(left[i], right[i]) - A[i];
26             if (height < 0)
27                 height = 0;
28             water += height;
29         }
30         
31         return water;
32     }
33 };

 

 1 struct Node
 2 {
 3     int val;
 4     int index;
 5     Node(){}
 6     Node(int v, int idx):val(v), index(idx){}
 7 };
 8 
 9 class Solution {
10 public:
11     int trap(int A[], int n) {
12         // Start typing your C/C++ solution below
13         // DO NOT write int main() function
14         stack<Node> s;
15         int sum = 0;
16         for(int i = 0; i < n; i++)
17             if (s.empty())
18                 s.push(Node(A[i], i));
19             else
20             {
21                 int height = 0;
22                 while(!s.empty())
23                 {
24                     Node node = s.top();
25                     if (node.val > A[i])
26                     {
27                         int width = i - node.index - 1;
28                         sum += A[i] * width - height * width;
29                         s.push(Node(A[i], i));
30                         break;
31                     }
32                     else
33                     {
34                         int width = i - node.index - 1;
35                         sum += node.val * width - height * width;
36                         height = node.val;
37                         s.pop();
38                     }
39                 }
40                 
41                 if (s.empty())
42                     s.push(Node(A[i], i));
43             }
44         
45         return sum;
46     }
47 };
posted @ 2012-11-25 19:26  chkkch  阅读(1126)  评论(0)  编辑  收藏  举报