LeetCode - Distinct Subsequences 题解

题目：

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.

解法一：搜索，超时

class Solution {public:    int C(int n, int m){        int ans = 1;        for(int i = n+1;i <= m; i++)            ans = ans * i;        for(int i = 1; i <= m-n; i++)            ans /= i;        return ans;    }    int numDistinct(string S, string T) {        if(T == "")            return 1;        int i = 0;        while(i < S.length() && S[i] != T[0])            i++;        if(i == S.length())            return 0;        int ns = 0, nt = 0;        while(i + ns < S.length() && S[i + ns] == S[i])            ns++;        while(nt < T.length() && T[0] == T[nt])            nt++;        int minN = min(ns, nt);        int ans = 0;        for(int j = 0; j <= minN; j++){            ans += C(j, ns) * numDistinct(S.substr(i + ns, S.length()), T.substr(j, T.length()));        }        return ans;    }};

解法二，动规

令f[i,j] 表示s的0..i里T0..j的distinct subsequence的个数，则

f[i,j]  = f[i-1,j] + f[i-1,j-1] * I {S[i] == T[j]},  I为指示函数

class Solution {public:    int numDistinct(string S, string T) {        vector<int> f(T.length(), 0), f0(T.length(), 0);        if(S[0] == T[0])            f0[0] = 1;        for(int i = 1; i < S.length(); i++){            f[0] = f0[0];            if(S[i] == T[0])                f[0] += 1;            for(int j = 1; j < T.length(); j++){                f[j] = f0[j];                if(S[i] == T[j])                    f[j] += f0[j-1];            }            f0 = f;        }        return f0[T.length() - 1];    }};