[leetcode]Distinct Subsequences @ Python
題意：
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.
解題思路：這道題使用動態規劃來解決。題的意思是：S的所有子串中，有多少子串是T。下面來看看狀態轉移方程。dp[i][j]表示S[0...i-1]中有多少子串是T[0...j-1]。
當S[i-1]=T[j-1]時：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]}
當S[i-1]!=T[j-1]時：dp[i][j]=dp[i-1][j-1]
初始化狀態如何確定呢：
dp[0][j]=0；因為：S是空串，則無論如何都不能包含非空的子串。這個初始狀態在初始化矩陣dp的時候就順帶包括了
dp[i][0]=1；因為：S[0...i-1]只有一個子串是空串。
代碼：
復制代碼
class Solution:
# @return an integer
def numDistinct(self, S, T):
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 range(1, len(S) + 1):
for j in range(1, len(T) + 1):
if S[i - 1] == T[j - 1]:
dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
else:
dp[i][j] = dp[i-1][j]
return dp[len(S)][len(T)]
