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