Distinct Subsequences

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.

用中文简要描述题目：在一个母串中，找出子串的个数（母串中允许删减字母，但不许颠倒顺序得到的字符串）

Java代码如下：

private int solution(String S, String T) {    if(S.length() < T.length() || T.length()*S.length() == 0) {        return 0;           } else if(S.equals(T)) {        return 1;           }    int[] count = new int[T.length()];          for(int indexS = 0; indexS < S.length(); indexS++) {                    char tempS = S.charAt(indexS);        for(int indexT = Math.min(indexS, T.length()-1); indexT >= 0;indexT--) {            if(tempS == T.charAt(indexT)) {                if(indexT == 0) {                    count[indexT] += 1;                } else {                    count[indexT] += count[indexT-1];                }            }           }                       }    return count[T.length()-1];}

思路：

1. 子串的个数决定于相同字符的取法？在特定范围内取法研究，比如在S = “rabbbit”, T = “rabbit”,T中的T.charAt(2,4)-“bb”,对应S中的S.charAt(2,5)-“bbb”,那么问题就可以归结为组合数问题C(3,2)
2. 组合数有如下特点：
   1 (n = 0)
   1 1 (n = 1)
   1 2 1 (n = 2)
   1 3 3 1 (n = 3)
   1 4 6 4 1 (n = 4)
   1 5 10 5 1 (n = 5)
   特点为：C(1, 1) = 1, C(n, 0) = 1, C(n, m) = C(n-1, m) + C(n-1, m-1)
3. 如果我们把上例中的结果用count[]数组来存，那么由于index0也有可能是组合数之一，所以count[0] += 1；(i>0)时，有count[i] += count[i-1]！