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
.
该题实际上是要求S到T只通过删除操作进行变换的方法数(通过从S中删去几个字符变成T的方法)
通过动归,对S前i位与T前j位的情况 考虑均新增一位会如何变化
class Solution { public: int numDistinct(string S, string T) { // Note: The Solution object is instantiated only once and is reused by each test case. vector<vector<int> > count; if(S.length()<T.length())return 0; count.resize(S.length()+1); for(int i=0;i<=S.length();i++){ count[i].resize( min(i,(int)T.length()) +1); count[i][0]=1; } for(int i=1;i<=S.length();i++) { if(count[i].size()==i+1){ int j=1; for(;j<count[i].size()-1;j++){ if(S[i-1]==T[j-1]){ count[i][j]=count[i-1][j-1]+count[i-1][j]; } else{ count[i][j]=count[i-1][j]; } } if(S[i-1]==T[j-1]){ count[i][j]=count[i-1][j-1]; } else{ count[i][j]=0; } } else{ int j=1; for(;j<count[i].size();j++){ if(S[i-1]==T[j-1]){ count[i][j]=count[i-1][j-1]+count[i-1][j]; } else{ count[i][j]=count[i-1][j]; } } } } return count[S.length()][T.length()]; } };