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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.
class Solution { public: int numDistinct(string s, string t) { int ls = s.length(), lt = t.length(), i, j; vector<vector<int>> dp(lt+1, vector<int>(ls+1, 0)); for(i = 0; i <= ls; i++) dp[0][i] = 1; for(i = 1; i <= lt; i++) { for(j = i; j <= ls; j++) { if(t[i-1] == s[j-1]) { dp[i][j] = dp[i-1][j-1]+dp[i][j-1]; } else { dp[i][j] = dp[i][j-1]; } } } return dp[lt][ls]; } };
注意：
“bbb”和“”的匹配结果为1。所以第一行都为1。
115. Distinct Subsequences *HARD* -- 字符串不连续匹配
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原文地址：http://www.cnblogs.com/argenbarbie/p/5448592.html
