题目：
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.
思路：
动态规划；关键是如何得到递推关系，可以这样想，设母串的长度为 j，子串的长度为 i，我们要求的就是长度为 i 的字串在长度为 j 的母串中出现的次数，设为 dp[i][j]，若母串的最后一个字符与子串的最后一个字符不同，则长度为 i 的子串在长度为 j 的母串中出现的次数就是母串的前 j - 1 个字符中子串出现的次数，即 dp[i][j] = dp[i][j - 1]，若母串的最后一个字符与子串的最后一个字符相同，那么除了前 j - 1 个字符出现字串的次数外，还要加上子串的前 i - 1 个字符在母串的前 j - 1个字符中出现的次数，即dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]。
package dp; public class DistinctSubsequences { public int numDistinct(String s, String t) { int slen = s.length(); int tlen = t.length(); if (tlen > slen) return 0; int[][] dp = new int[slen + 1][tlen + 1]; for (int i = 0; i <= slen; ++i) { for (int j = 0; j <= tlen; ++j) { if (j == 0) { dp[i][j] = 1; continue; } if (j > i) { dp[i][j] = 0; continue; } if (s.charAt(i - 1) == t.charAt(j - 1)) dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1]; else dp[i][j] = dp[i - 1][j]; } } return dp[slen][tlen]; } public static void main(String[] args) { // TODO Auto-generated method stub DistinctSubsequences d = new DistinctSubsequences(); System.out.println(d.numDistinct("rabbbit", "rabbit")); } }
LeetCode - Distinct Subsequences
原文：http://www.cnblogs.com/shuaiwhu/p/5127175.html