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"));    }}