LeetCode-Distinct Subsequences

典型的dp，

最开始写了个递归的，过不了大数据，

 1 class Solution {
 2 public:
 3     int numDistinct(string S, string T) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function
 6         int m = S.length();
 7         int n = T.length();
 8         return getDistinct(S, T, m - 1, n - 1);
 9     }
10     int getDistinct(string &S, string &T, int i, int j) {
11         if (j < 0) {
12             return 1;
13         }
14         if (i < 0) {
15             return 0;
16         }
17         if (j > i) {
18             return 0;
19         }
20         if (S[i] == T[j]) {
21             return (getDistinct(S, T, i - 1, j - 1) + getDistinct(S, T, i - 1, j));
22         }
23         return getDistinct(S, T, i - 1, j);
24     }
25 };

哎，看来做DP的题目的第一选择就应该是直接用数组保存最优解。

 1 class Solution {
 2 public:
 3     int numDistinct(string S, string T) {
 4         int N = S.size(), M = T.size();
 5         int dp[M+1][N+1];
 6         dp[0][0] = 1;
 7         for (int j = 1; j <= N; ++j)
 8             dp[0][j] = 1;
 9         for (int i = 1; i <= M; ++i)
10             dp[i][0] = 0;
11 
12         for (int i = 1; i <= M; ++i)
13             for (int j = 1; j <= N; ++j)
14                 if (S[j-1] == T[i-1])
15                     dp[i][j] = dp[i][j-1] + dp[i-1][j-1];
16                 else
17                     dp[i][j] = dp[i][j-1];
18 
19         return dp[M][N];
20     }
21 };

 

posted @ 2013-09-01 10:17  Exio  阅读(117)  评论(0)  编辑  收藏  举报