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

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"

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

class Solution(object):
    def numDistinct(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: int
        """
        # 關鍵是定義各個狀態，找出狀態轉移方程
        # dp[i][j]表示字符串s[0~i-1] 中有多少個t[0~j-1].
 

        dp = [[0 for col in range(len(t) + 1)] for row in range(len(s) + 1)]
        if len(t) == 0 or len(s) == 0:
            return 0
        for i in range(len(s) + 1):     # 給第一列邊界值賦值,此時為s[0]，即為空時
            dp[i][0] = 1
        for i in range(1,len(s) + 1):
            for j in range(1, len(t) + 1):
                
 
if s[i - 1] == t[j - 1]:  # 因為比dp的維度要小1，所以為i-1,j-1
                    #如果S的第i個字符和T的第j個字符相同，那麽所有dp[i-1][j-1]中滿足的結果都會成為新的滿足的序列
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
                else:
                    #假設S的第i個字符和T的第j個字符不相同，那麽就意味著dp[i][j]的值跟res[i-1][j]是一樣
                    dp[i][j] = dp[i - 1][j]
        return dp[len(s)][len(t)]
            
        
        

Leetcode -- 115. Distinct Subsequences