LeetCode-115-Distinct Subsequences

算法描述：

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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).

Example 1:

Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^

Example 2:

Input: S = "babgbag", T = "bag"
Output: 5
Explanation:

As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)

babgbag
^^ ^
babgbag
^^    ^
babgbag
^    ^^
babgbag
  ^  ^^
babgbag
    ^^^

解题思路：动态规划题。如果字符s[i-1]和t[j-1]相等，则结果序列分为两种情况，包含当前字符的序列和不包含当前字符的序列。如果不相等，则只有一种情况，即不包含当前字符的序列。空串是任何序列的子序列。

递推公式：

dp[i][j] = dp[i-1][j-1] + dp[i-1][j] if(s[i-1]==t[j-1]);

dp[i][j] = dp[i-1][j] if(s[i-1]!=t[j-1]);

    int numDistinct(string s, string t) {
        if(s.size()==0 || t.size()==0) return 0;
        vector<vector<long long>> dp(s.size()+1,vector<long long>(t.size()+1,0));
        
        for(int i =0; i <= s.size(); i++) dp[i][0] = 1;
        for(int i =1; i <= s.size(); i++){
            for(int j=1; j <= t.size(); 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[s.size()][t.size()];
    }