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"
Return3.
思路：
1. 初始化一个矩阵number[i][j]用来记录字符串T的前j个字符出现在字符串S的前i个字符的次数，当j=0时，令number[i][j]=1；
2. 当S的第i个字符与T的第j个字符不同时，则说明S的第i个字符对number[i][j]没有影响，即number[i][j]=number[i-1][j]；
3. 当S的第i个字符与T的第j个字符不同时，则说明S的第i个字符对number[i][j]有影响，number[i][j]除了要算上原来的number[i-1][j]，还要算上新的可能性，即number[i-1][j-1].
例子
| 0 | r | a | b | b | i | t | |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| r | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| a | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| b | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
| b | 1 | 1 | 1 | 2 | 1 | 0 | 0 |
| b | 1 | 1 | 1 | 3 | 3 | 0 | 0 |
| i | 1 | 1 | 1 | 3 | 3 | 3 | 0 |
| t | 1 | 1 | 1 | 3 | 3 | 3 | 3 |
代码：
1 public static int result(String str1, String str2){ 2 int len1 = str1.length(), len2 = str2.length(); 3 int[][] res = new int[len1+1][len2+1]; 4 for(int i=0;i<len1+1;i++){ 5 res[i][0] = 1; 6 } 7 for(int i=1;i<len1+1;i++){ 8 for(int j=1;j<len2+1;j++){ 9 if(str1.charAt(i-1)!=str2.charAt(j-1)) 10 res[i][j] = res[i-1][j]; 11 else 12 res[i][j] = res[i-1][j]+res[i-1][j-1]; 13 } 14 15 } 16 return res[len1][len2]; 17 }