leetcode 131. Palindrome Partitioning

题目链接：leetcode

思路：
回溯

class Solution:
    def partition(self, s):
        """
        :type s: str
        :rtype: List[List[str]]
        """
        self.s =s

        def dfs(s,res,tmp):
            if not s:
                res.append(tmp[:])
                return
            for i in range(1,len(s)+1):
                stmp = s[:i]
                if stmp == stmp[::-1]:
                    #这行代码等价于
                    #tmp.append(stmp)
                    #dfs(s[i:],res,tmp)
                    #tmp.pop()
                    dfs(s[i:],res,tmp+[stmp])
        res=[]
        dfs(self.s,res,[])
        return res