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
.
方法:dp
a[i][j]表示T中前j个在S的前i个出现的个数。
则
if S[i] != T[j]
a[i][j] = a[i-1][j],相当于如果S中前i-1个可以表示T中前j个的话,那么多一个S[i]也不会影响。
if S[i] == T[j]
a[i][j] = a[i-1][j] + a[i-1][j-1]
class Solution { public: int numDistinct(string S, string T) { int* a = new int[T.size()]; memset(a, 0, sizeof(int) * T.size()); for (int i = 0; i < S.size(); i++) { for (int j = T.size()-1; j >= 0; j--) { if (S[i] == T[j]) { if (j==0) a[j] += 1; else a[j] += a[j-1]; } } } return a[T.size()-1]; } };