LeetCode每日一题：不同的子序列数量

问题描述

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"
Return3.

问题分析

这题是让我们通过删除字符来实现从S转化成T，并返回有多少种方法，典型的动态规划DP题。

• 定义dp[i][j]表示S的前i个字符转化成T的前j个字符的方法数
• 若S[i]==T[j],dp[i][j]=dp[i-1][j-1]+dp[i-1][j]+dp[i][j]
  表示的意思是，分三种情况：
  • 第i个字符和第j个字符不进行匹配
  • 用S的前i-1个字符来匹配T的前j个字符
  • 用S的前i个字符来匹配T的前j个字符
• 若S[i]！=T[j]，则无法匹配，dp[i][j]=dp[i-1][j-1]

代码实现

public int numDistinct(String S, String T) {
        if (S == null || T == null) return 0;
        if (T.length() > S.length()) return 0;
        int[][] dp = new int[S.length() + 1][T.length() + 1];
        dp[0][0] = 1;
        for (int i = 1; i <= S.length(); i++) {
            dp[i][0] = 1;
        }
        for (int i = 1; i <= S.length(); i++) {
            for (int j = 1; j <= T.length(); j++) {
                if (S.charAt(i - 1) != T.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j];
                } else if (S.charAt(i - 1) == T.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] + dp[i][j];
                }
            }
        }
        return dp[S.length()][T.length()];
    }