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palindrome-partitioning.py 1.59 KB
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Allen Liu 提交于 2018-10-13 13:24 +08:00 . update
# Time: O(n^2 ~ 2^n)
# Space: O(n^2)
class Solution(object):
# @param s, a string
# @return a list of lists of string
def partition(self, s):
n = len(s)
is_palindrome = [[0 for j in xrange(n)] for i in xrange(n)]
for i in reversed(xrange(0, n)):
for j in xrange(i, n):
is_palindrome[i][j] = s[i] == s[j] and ((j - i < 2 ) or is_palindrome[i + 1][j - 1])
sub_partition = [[] for i in xrange(n)]
for i in reversed(xrange(n)):
for j in xrange(i, n):
if is_palindrome[i][j]:
if j + 1 < n:
for p in sub_partition[j + 1]:
sub_partition[i].append([s[i:j + 1]] + p)
else:
sub_partition[i].append([s[i:j + 1]])
return sub_partition[0]
# Time: O(2^n)
# Space: O(n)
# recursive solution
class Solution2(object):
# @param s, a string
# @return a list of lists of string
def partition(self, s):
result = []
self.partitionRecu(result, [], s, 0)
return result
def partitionRecu(self, result, cur, s, i):
if i == len(s):
result.append(list(cur))
else:
for j in xrange(i, len(s)):
if self.isPalindrome(s[i: j + 1]):
cur.append(s[i: j + 1])
self.partitionRecu(result, cur, s, j + 1)
cur.pop()
def isPalindrome(self, s):
for i in xrange(len(s) / 2):
if s[i] != s[-(i + 1)]:
return False
return True
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