原题地址：https://oj.leetcode.com/problems/distinct-subsequences/

题意：

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.

解题思路：这道题使用动态规划来解决。题的意思是：S的所有子串中，有多少子串是T。下面来看看状态转移方程。dp[i][j]表示S[0...i-1]中有多少子串是T[0...j-1]。

　　　　　当S[i-1]=T[j-1]时：dp[i][j]=dp[i-1][j-1]+dp[i-1][j]；S[0...i-1]中有多少子串是T[0...j-1]包含：{S[0...i-2]中有多少子串是T[0...j-2]}+{S[0...i-2]中有多少子串是T[0...j-1]}

　　　　   当S[i-1]!=T[j-1]时：dp[i][j]=dp[i-1][j-1]

　　　　   那么初始化状态如何确定呢：dp[i][0]=1；S[0...i-1]只有一个子串是空串。

代码：

class Solution:
    # @return an integer
    # @dp
    # dp[i][j] means how many first j of T is sub of first i of S.
    def numDistinct(self, S, T):
        dp = [[0 for i in range(len(T)+1)] for j in range(len(S)+1)]
        for j in range(len(S)+1):
            dp[j][0] = 1
        for i in range(1, len(S)+1):
            for j in range(1, min(i+1, len(T)+1)):
                if S[i-1] == T[j-1]:
                    dp[i][j] = dp[i-1][j] + dp[i-1][j-1]
                else:
                    dp[i][j] = dp[i-1][j]
        return dp[len(S)][len(T)]