leetcode 115. Distinct Subsequences-优快云博客

博客围绕给定两个字符串S和T，计算S中等于T的不同子序列数量展开。指出这是典型的动态规划问题，分析了i、j从1开始时i、j等于0的处理难点，还阐述了dp[i][j]的含义及递推关系。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
    ^^^

思路：一道典型动态规划的题。

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

可以看出i，j都要从1开始，所以如何处理i，j等于0的问题就是一个小难点。

dp[i][j]是s的前i个字符可以形成的包含t的前j字符的子串的个数，如果s[i-1]==t[j-1],则dp[i][j] 的个数等于s中前i-1个字符可以形成的包含t中前j-1个字符的子串个数+s前i个字符可以形成的包含t的前j个字符的子串个数。

所以当j==1，如果s[i-1]==t[j-1],dp[i][j] = dp[i-1][j] + 1;

所以将dp[i][0] = 1;

class Solution {
public:
    int numDistinct(string s, string t) 
    {
        int len1 = s.size();
        int len2 = t.size();
        vector<long> dp(len2+1,0);
        dp[0] = 1;
        for(int i = 1; i <= len1; ++i)
        {
            for(int j = len2; j >= 1; --j)
            {
                if(s[i-1] == t[j-1]) dp[j] = dp[j] + dp[j-1];
            }
        }
        return dp[len2];
    }
};