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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.
这一题采用的是dp的思路。设置dp[i][j]表示字符串t[0->j)在字符串s[0->i)中出现的的数目。初始化设置dp[i][0] i∈[0,s.length() ),表示字符串s[0->i)变为空字符串只有一种方法。对于i∈[1,s.length()] j∈[1,t.length()],如果s[i-1]==t[j-1],那么dp[i][j]=dp[i-1][j-1]+dp[i-1][j],表示取s[i-1]对应t[j-1],或者不取s[i-1]对应t[j-1]的所有子串之和。如果如果s[i-1]！=t[j-1]，那么不能用s[i-1]对应t[j-1],dp[i][j]只能是dp[i-1][j]。
class Solution { public: int numDistinct(string s, string t) { vector<vector<int>>dp(s.length()+1,vector<int>(t.length()+1)); for(int i=0;i<s.length();i++) dp[i][0]=1; for(int i=1;i<=s.length();i++){ for(int j=1;j<=t.length();j++){ dp[i][j]=dp[i-1][j]; if(s[i-1]==t[j-1]){ dp[i][j]+=dp[i-1][j-1]; } } } return dp[s.length()][t.length()]; } };
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原文地址：http://www.cnblogs.com/zhoudayang/p/5045578.html
