LeetCode OJ:Distinct Subsequences

Distinct Subsequences

 

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.

算法思想：直接深度递归超时，递归+纪录路径可AC

class Solution{public:int dfs(vector<vector<int>> &dp,string &S,string &T,int s,int t){if(t==T.size())return 1;if(dp[t][s]!=-1)return dp[t][s];int sum=0;for(int i=s;i<S.size();i++){if(T[t]==S[i]){sum+=dfs(dp,S,T,i+1,t+1);}}dp[t][s]=sum;return sum;}int numDistinct(string S,string T){vector<vector<int>> dp(T.size());for(int i=0;i<T.size();i++){dp[i].assign(S.size()+1,-1);}return dfs(dp,S,T,0,0);}};

动态规划：

f(i,j)表示T[0,j]在S[0,i]里出现的次数，

若S[i]==T[j]

f(i,j)=f(i-1,j)+f(i-1,j-1);

else f(i,j)=f(i-1,j)

class Solution {public:    int numDistinct(string S, string T) {        vector<int> f(T.size()+1);        f[0]=1;        for(int i=0;i<S.size();++i){            for(int j=T.size()-1;j>=0;--j){                f[j+1]+=S[i]==T[j]?f[j]:0;            }        }                return f[T.size()];    }};