标签:style color io os ar for sp c on
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
.
//这里可以用dfs,但是,TLE.所以,采用dp来解决. //假设dp[i][j]的状态表示为字符串i变换到字符串j的方法的数量,那么,动态转移方程为: // 1) S[i-1] == T[j-1] --> dp[i][j] = dp[i-1][j-1] + dp[i-1][j] // 2) S[i-1] != T[j-1] --> dp[i][j] = dp[i-1][j]. class Solution { public: int numDistinct(std::string S, std::string T) { int n = S.size(),m = T.size(); std::vector<std::vector<int>> dp(n+1,std::vector<int>(m+1,0)); for (int i = 0; i < n+1; i++) { dp[i][0] = 1; } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if(S[i-1] == T[j-1]) dp[i][j] = dp[i-1][j-1] + dp[i-1][j]; else dp[i][j] = dp[i-1][j]; } } return dp[n][m]; } };
leetcode - Distinct Subsequences
标签:style color io os ar for sp c on
原文地址:http://blog.csdn.net/akibatakuya/article/details/39736237