118. Distinct Subsequences/115. Distinct Subsequences
- 本题难度: Medium/Hard
- Topic: Dynamic Programming
Description
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" is a subsequence of "ABCDE" while "AEC" is not).
Example
Given S = "rabbbit", T = "rabbit", return 3.
Challenge
Do it in O(n^2) time and O(n) memory.
O(n^2) memory is also acceptable if you do not know how to optimize memory.
我的代码
class Solution:
"""
@param S: A string
@param T: A string
@return: Count the number of distinct subsequences
"""
def numDistinct(self, S, T):
# write your code here
ls = len(S)
lt = len(T)
if lt == 0:
return 1
res = [0 for _ in range(ls)]
for i in range(ls):
if S[i] == T[0]:
res[i] = 1
for i in range(1,lt):
pre_sum = 0
for j in range(ls):
now = res[j]
if S[j] == T[i]:
res[j] = pre_sum
else:
res[j] = 0 #忘记更新了
pre_sum += now
return sum(res)
思路
一开始想模仿119. Edit Distance/72. Edit Distance的思路，初始设定空间复杂度为O(n^2)
“dp[i][j] 对S中S[:i+1]和T[:j+1]有多少种情况，当S[i]==T[j]时，如“rabb”和“rab”等于“r”和“ra”的情况+1”
但是突然发现新增加的元素可能在前面就会有重复，所以应该分成两部分来看，即“ra”在S各个位置的情况+各个位置之后“b”的情况。
然后我模拟了一下res的变化，发现res的变化很有规律，可以in-place更新。所以空间复杂度变成了O(n)
S = "rabbbit", T = "rabbit",
r a b b b i t
r 1 0 0 0 0 0 0 1
ra 0 1 0 0 0 0 0 1
rab 0 0 1 1 1 0 0 3
rabb 0 0 0 1 2 0 0 3
rabbi 0 0 0 0 0 3 0 3
rabbit 0 0 0 0 0 0 3 3- 时间复杂度: O(mn)
- 空间复杂度: O(n)
- 出错
忘记更新在S[j] != T[i]时的res[i][j]=0