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  • 115. Distinct Subsequences *HARD* -- 字符串不连续匹配

    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.

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

    注意:

    “bbb”和“”的匹配结果为1。所以第一行都为1。

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  • 原文地址:https://www.cnblogs.com/argenbarbie/p/5448592.html
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