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.

Solution

• 把s删掉某个或者几个，得到t
• dp[i][j]:表示S[0…i-1]中有多少子串是T[0…j-1]

dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)

dp[i][j] = dp[i][j - 1].就是说，假设S已经匹配了j - 1个字符，得到匹配个数为dp[i][j - 1].现在无论S[j]是不是和T[i]匹配，匹配的个数至少是dp[i][j - 1]。除此之外，当S[j]和T[i]相等时，我们可以让S[j]和T[i]匹配，然后让S[j - 1]和T[i - 1]去匹配。所以递推关系

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 class Solution:
    # @param S, T: Two string.
    # @return: Count the number of distinct subsequences
    def numDistinct(self, S, T):
        # write your code here
        dp = [[0 for j in range(len(T) + 1)] for i in range(len(S) + 1)]
        for i in range(len(S) + 1):
              dp[i][0] = 1
        for i in xrange(len(S)):
            for j in xrange(len(T)):
                if S[i] == T[j]:
                    dp[i+1][j+1] = dp[i][j+1] + dp[i][j]
                else:
                    dp[i+1][j+1] = dp[i][j + 1]
        return dp[len(S)][len(T)]