Leetcode 115 Distinct Subsequences

题目来源：leetcode

Given a string S and a string T, count the number of distinct subsequences ofS which equalsT.

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：来自leetcode  dp二维数组记录定义t[0...i] s[0...j]所包含字串的个数 

     动态规划转移方程：dp[i+1][j+1]=dp[i][j]+dp[i+1][j]   当s与t相应位置字符串想的时候。数量为不计算这两个相等的字符dp[i][j]加上不包括t的该位置字符之前相等过的字符dp[i+1][j]

class Solution {    public int numDistinct(String s, String t) {        int [][]dp=new int[t.length()+1][s.length()+1];//定义t[0...i] s[0...j]所包含字串的个数            for(int i=0;i<=s.length();i++)                dp[0][i]=1;//空字符串是所有字符串的子串        for(int i=1;i<t.length();i++)            dp[i][0]=0;//若s为空是0        for(int i=0;i<t.length();i++)            for(int j=0;j<s.length();j++)                if(s.charAt(j)==t.charAt(i)){                    dp[i+1][j+1]=dp[i][j]+dp[i+1][j];                }            else dp[i+1][j+1]=dp[i+1][j];//如果不相等，跟前一个s的字符一样     return dp[t.length()][s.length()];       }}

另一较多的思想：通过递归不断比较T的起始字符和S中想的的的位置递归继续求下一个是否存在，一直到最后若存在则记为一个字串 ，最后回溯总的个数