LeetCode-Trapping Rain Water

又是这种需要仔细考虑的题目，

很容易就会漏掉一些情况，哎，感冒了，

完全不在状态，调了好几个小时，

写代码的能力实在是不行，总感觉写出来的代码不够优雅，

 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         int res = 0;
 7         if (A == NULL) {
 8             return res;
 9         }
10         int i = 1;
11         int b = 0;
12         int low;
13         while (i < n && A[i] >= A[i - 1]) {
14             b = i;
15             ++i;
16         }
17         while (i < n) {
18             while (i < n && A[i] < A[i - 1]) {
19                 ++i;
20             }
21             int k = i - 1;
22             int j = k - 1;
23             low = A[k];
24             while (i < n && A[i] >= A[i - 1]) {
25                 if (A[i] > A[i - 1]) {
26                     if (j >= b) {
27                         while (j >= b && A[i] >= A[j]) {
28                             int len = i - j - 1;
29                             int h = A[j] - low;
30                             low = A[j];
31                             res += len * h;
32                             while (j >= b && A[j] <= low) 
33                                 --j;
34                         }
35                         if (j >= b) {
36                             int len = i - j - 1;
37                             int h = A[i] - low;
38                             low = A[i];
39                             res += len * h;
40                         }
41                     }
42                     ++i;
43                     continue;
44                 }
45                 ++i;
46             }
47             if (A[b] <= A[i - 1]) {
48                 b = i - 1;
49             }
50             ++i;
51         }
52         return res;
53     }
54 };

大神们的方法，超级简单，找左右两边的最大值即可；

 1 class Solution {
 2 public:
 3     int trap(int A[], int n) {
 4         if (n == 0) return 0;
 5         int left[n];
 6         int right[n];
 7 
 8         int lmax = A[0];
 9         for (int i = 0; i < n; ++i)
10         {
11             lmax = max(lmax, A[i]);
12             left[i] = lmax;
13         }
14 
15         int rmax = A[n-1];
16         for (int i = n - 1; i >= 0; --i)
17         {
18             rmax = max(rmax, A[i]);
19             right[i] = rmax;
20         }
21 
22         int res = 0;
23         for (int i = 0; i < n; ++i)
24             res += min(left[i], right[i]) - A[i];
25 
26         return res;
27     }
28 };

 

posted @ 2013-09-20 01:00  Exio  阅读(151)  评论(0)  编辑  收藏  举报