https://leetcode.com/problems/palindrome-partitioning/
分割问题，找分割点。这里可以有多个分割点，本质就是选取分割点的子集，能够满足palindrome的条件即可。运用普通的dfs就行
class Solution(object):
def checkPalindrome(self, s):
if len(s) == 0: return True
i,j = 0, len(s) - 1
while i<=j:
if s[i] != s[j]:
return False
i += 1
j -= 1
return True
def dfs(self, candidates, start, end, subres, res):
if start > end:
res.append(subres[:])
return
else:
for i in xrange(start, end + 1):#对于start到end的candidates, 其分割点应该是包括i从start到end每个点，candidates[:i+1]即为被分割的部分，包括整个candidates
if self.checkPalindrome(candidates[start: i + 1]):
self.dfs(candidates, i + 1, end, subres + [candidates[start: i + 1]], res)
def partition(self, s):
"""
:type s: str
:rtype: List[List[str]]
"""
res = []
self.dfs(s, 0, len(s)-1,[],res)
return res自己复习重写code
class Solution(object):
def isP(self, s):
i,j = 0, len(s) - 1
while i < j:
if s[i] != s[j]:
return False
i += 1
j -= 1
return True
def dfs(self, s, subres, res):
if len(s) == 0:
res.append(subres)
return
for i in xrange(len(s)):
if self.isP(s[:i+1]):
self.dfs(s[i+1:], subres + [s[:i+1]], res)
def partition(self, s):
"""
:type s: str
:rtype: List[List[str]]
"""
if not s: return []
res = []
self.dfs(s, [], res)
return res
