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rotate-string.py 2.53 KB
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Allen Liu 提交于 2018-10-13 01:56 +08:00 . add complexity
# Time: O(n)
# Space: O(1)
class Solution(object):
def rotateString(self, A, B):
"""
:type A: str
:type B: str
:rtype: bool
"""
def check(index):
return all(A[(i+index) % len(A)] == c
for i, c in enumerate(B))
if len(A) != len(B):
return False
M, p = 10**9+7, 113
p_inv = pow(p, M-2, M)
b_hash, power = 0, 1
for c in B:
b_hash += power * ord(c)
b_hash %= M
power = (power*p) % M
a_hash, power = 0, 1
for i in xrange(len(B)):
a_hash += power * ord(A[i%len(A)])
a_hash %= M
power = (power*p) % M
if a_hash == b_hash and check(0): return True
power = (power*p_inv) % M
for i in xrange(len(B), 2*len(A)):
a_hash = (a_hash-ord(A[(i-len(B))%len(A)])) * p_inv
a_hash += power * ord(A[i%len(A)])
a_hash %= M
if a_hash == b_hash and check(i-len(B)+1):
return True
return False
# Time: O(n)
# Space: O(n)
# KMP algorithm
class Solution2(object):
def rotateString(self, A, B):
"""
:type A: str
:type B: str
:rtype: bool
"""
def strStr(haystack, needle):
def KMP(text, pattern):
prefix = getPrefix(pattern)
j = -1
for i in xrange(len(text)):
while j > -1 and pattern[j + 1] != text[i]:
j = prefix[j]
if pattern[j + 1] == text[i]:
j += 1
if j == len(pattern) - 1:
return i - j
return -1
def getPrefix(pattern):
prefix = [-1] * len(pattern)
j = -1
for i in xrange(1, len(pattern)):
while j > -1 and pattern[j + 1] != pattern[i]:
j = prefix[j]
if pattern[j + 1] == pattern[i]:
j += 1
prefix[i] = j
return prefix
if not needle:
return 0
return KMP(haystack, needle)
if len(A) != len(B):
return False
return strStr(A*2, B) != -1
# Time: O(n^2)
# Space: O(n)
class Solution3(object):
def rotateString(self, A, B):
"""
:type A: str
:type B: str
:rtype: bool
"""
return len(A) == len(B) and B in A*2
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