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Distinct Subsequences
-----QUESTION-----
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
.
-----SOLUTION-----
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]; } };
以下做法在Judge Large时,time limit exceeded
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 ++; } };
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