LeetCode之“动态规划”：Distinct Subsequences

　　题目链接

　　题目要求：

　　Given a string S and a string T, count the number of distinct subsequences of T in S.

　　A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

　　Here is an example:
　　S = "rabbbit", T = "rabbit"

　　Return 3.

　　该题解析参考自LeetCode题解。

　　设状态为dp[i][j]，表示T[0, j]在S[0, i]里出现的次数。首先，无论S[i]和T[j]是否相等，若不使用S[i]，则dp[i][j]=dp[i-1][j]；若S[i]=T[j]，则可以使用S[i]，此时dp[i][j]=dp[i-1][j]+dp[i-1][j-1]。

　　代码如下：

 1 class Solution {
 2 public:
 3     int numDistinct(string s, string t) {
 4         int szS = s.size();
 5         int szT = t.size();
 6         if(szS < szT)
 7             return 0;
 8
 9         vector<vector<int> > dp(szS + 1, vector<int>(szT + 1, 0));
10         for(int i = 0; i < szS + 1; i++)
11             dp[i][0] = 1;
12
13         for(int i = 1; i < szS + 1; i++)
14             for(int j = 1; j < szT + 1; j++)
15             {
16                 if(s[i-1] != t[j-1])
17                     dp[i][j] = dp[i-1][j];
18                 else
19                     dp[i][j] = dp[i-1][j] + dp[i-1][j-1];
20             }
21
22         return dp[szS][szT];
23     }
24 };