LeetCode: Distinct Subsequences

一开始dfs没过large，网上看了别人的dp，哎，这题比较难。。

 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         vector<vector<int>> f(S.size()+1, vector<int>(T.size()+1));
 7         for (int i = 0; i <= S.size(); i++) f[i][0] = 0;
 8         for (int i = 0; i <= T.size(); i++) f[0][i] = 0;
 9         for (int i = 1; i <= S.size(); i++) {
10             if (S[i-1] == T[0]) f[i][1] = f[i-1][1]+1;
11             else f[i][1] = f[i-1][1];
12         }
13         for (int i = 2; i <= S.size(); i++) {
14             for (int j = 2; j <= T.size(); j++) {
15                 if (S[i-1] == T[j-1]) f[i][j] = f[i-1][j-1]+f[i-1][j];
16                 else f[i][j] = f[i-1][j];
17             }
18         }
19         return f[S.size()][T.size()];
20     }
21 };

 1 public class Solution {
 2     public int NumDistinct(string s, string t) {
 3         int[,] f = new int[s.Length+1, t.Length+1];
 4         for (int i = 1; i <= s.Length; i++) {
 5             if (s[i-1] == t[0]) f[i, 1] = f[i-1, 1] + 1;
 6             else f[i, 1] = f[i-1, 1];
 7         }
 8         for (int i = 2; i <= s.Length; i++) {
 9             for (int j = 2; j <= t.Length; j++) {
10                 if (s[i-1] == t[j-1]) f[i, j] = f[i-1, j-1] + f[i-1, j];
11                 else f[i, j] = f[i-1, j];
12             }
13         }
14         return f[s.Length, t.Length];
15     }
16 }

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posted @ 2013-03-20 17:10 ying_vincent 阅读(196) 评论(0) 收藏举报

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