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()];
}
};