LeetCode——Distinct Subsequences(115)

最新推荐文章于 2022-07-17 21:32:36 发布

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本文探讨了如何使用动态规划解决字符串子序列匹配问题，并详细解释了算法背后的逻辑与实现步骤。

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
    ^^^

分析：

这题就不装了，没想出来，看的牛客网的解析，很厉害，怎么想到的....想了半天没想懂二维的dp表怎么建。

https://www.nowcoder.com/questionTerminal/ed2923e49d3d495f8321aa46ade9f873

代码：

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