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 tLen = T.length();
int sLen = S.length();
if(sLen == 0) return 0;
vector<vector<int>> dp(sLen,vector<int>(tLen));
int i=0;
int j=0;
if(S[0]==T[0]) dp[0][0] = 1;
else dp[0][0]=0;
for(i = 1; i < tLen; i++)
dp[0][i]=0;
for(i = 1 ; i < sLen ;i++)
if(S[i]==T[0])
dp[i][0] = dp[i-1][0] + 1;
else
dp[i][0] = dp[i-1][0];
for(int i = 1; i < sLen ; i++)
for(int j = 1; j < tLen ; j++)
if(S[i]==T[j])
dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
else
dp[i][j] = dp[i-1][j];
return dp[sLen-1][tLen-1];
}
};class Solution {
public:
int numDistinct(string S, string T) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int result = 0;
solve(S,T,0,0,result);
return result;
}
void solve(const string & S, const string & T, int sPos, int tPos, int & result)
{
int tLen = T.length();
int sLen = S.length();
for (int i = tPos; i< tLen; i++)
{
if(sPos >= sLen)
return;
while(S[sPos]!=T[i])
{
sPos++;
if(sPos >= sLen)
return;
}
sPos++;
solve(S,T,sPos,i,result);
}
result ++;
}
};Distinct Subsequences,布布扣,bubuko.com
原文：http://blog.csdn.net/joannae_hu/article/details/38128469