LeetCode 115 Distinct Subsequences (子序列计数 dp)

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本文介绍了一种计算字符串T在字符串S中不同子序列数量的方法。通过动态规划算法实现，利用二维数组dp记录中间结果，避免重复计算。文章提供了一个具体的例子进行说明，并附带了完整的Java代码实现。

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.

题目链接：https://leetcode.com/problems/distinct-subsequences/

题目分析：设dp[i][j]表示T的前缀0~(i-1)在S的前缀0~(j-1)中出现的次数，首先有dp[i][j] = dp[i][j - 1]，若t[i] == s[j]则dp[i][j] += dp[i - 1][j - 1]

S="abcab"，T="ab"

       0 1 2 3 4 5
          a b c a b
0    1 1 1 1 1 1
1 a 0 1 1 1 2 2
2 b 0 0 1 1 1 3

public class Solution {
    public int numDistinct(String s, String t) {
        int lens = s.length();
        int lent = t.length();
        if (lens < lent) {
            return 0;
        }
        int[][] dp = new int[lent + 1][lens + 1];
        for (int j = 0; j < lens; j ++) {
            dp[0][j] = 1;
        }
        for (int i = 0; i < lent; i ++) {
            for (int j = 0; j < lens; j ++) {
                dp[i + 1][j + 1] = dp[i + 1][j];
                if (t.charAt(i) == s.charAt(j)) {
                    dp[i + 1][j + 1] += dp[i][j];
                }
            }
        }
        return dp[lent][lens];
    }
}