*leetcode #115 in cpp

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本文介绍了一种算法，用于计算一个字符串作为另一个字符串子序列出现的次数。提供了两种解决方案：一种是使用递归加剪枝的方法；另一种是采用动态规划算法。

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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.

Code:Recursive with pruning.

class Solution {
public:
    int numDistinct(string s, string t) {
        vector<vector<int>> mem(s.length(), vector<int>(t.length(),-1)); 
        if(t == "") return 1; 
        return findNumDistinct(s,t,0, 0, mem);
    }
    
    int findNumDistinct(string &s, string &t, int ps, int pt,vector<vector<int>> &mem){
        if(pt >= t.length()) return 1; 
        if(ps >= s.length()) return 0;
        if(mem[ps][pt] != -1) return mem[ps][pt];
        int res = 0;
        for(int i = ps; i < s.length(); i ++){
            if(s[i] == t[pt]){
                res += findNumDistinct(s,t,i+1, pt+1,mem);
            }
        }
        mem[ps][pt] = res;
        return res;
    }
};

Code: DP

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.length();
        int n = t.length(); 
        vector<vector<int>> dp(m+1,vector<int>(n+1,0));
        dp[0][0] = 1; 
        for(int i = 1; i <= m; i ++){
            dp[i][0] = 1;
        }
        for(int i = 1; i <= n; i ++){
            dp[0][i] = 0;
        }
        for(int i = 1; i <= m; i ++){
            for(int j = 1; j <= n; j++){
                dp[i][j] = dp[i-1][j];
                if(s[i-1] == t[j-1]){
                    dp[i][j] += dp[i-1][j-1];
                }
            }
        }
        
        
        return dp[m][n];
        
    }
    
};