Leetcode-115. Distinct Subsequences

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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
    ^^^

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Submissions

287,652

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

此题为hard难度，其解法类似与算法课上的找最长字串，而这里是找在S中有几个T子串。实际上就是用DP方法去维护一个二维数组，这个数组的的dp[i][j]就是对于t[0~i]和s[0~j]的此题的解。（也就是在此时s中有多少个t子串）。然后通过动态规划的化解可以知道，dp[i][j]=dp[i][j-1]（如果两个串最后一个值不相等)，dp[i][j]=dp[i-1][j-1]+dp[i][j-1]（如果最后一个值相等）再根据初始值和边值条件，对数组进行遍历赋值，以得最后的解。