python--lintcode136. 分割回文串

描述

给定一个字符串s，将s分割成一些子串，使每个子串都是回文串。

返回s所有可能的回文串分割方案。

您在真实的面试中是否遇到过这个题？  

样例

给出 s = "aab"，返回

[
  ["aa", "b"],
  ["a", "a", "b"]
]

这一题我本来的思路是，首先得到所有的排列，即在哪个位置可以分割，然后再分割字符串，再将分割得到的每一个子字符串进行判断是否是回文。思路是挺清晰的，但是结果竟然超时了：

class Solution:
    """
    @param: s: A string
    @return: A list of lists of string
    """
    def partition(self, s):
        # write your code here
        def search(totalyString,subString,index):  #得到所有排列组合
            if(index==len(totalyString)-1):
                self.result.append(subString)
                return
            search(totalyString,subString+[index],index+1)
            search(totalyString,subString,index+1)

            return


        def ishuiwen(substring):       #判断是否是回文字符串
            begin,end=0,len(substring)-1
            while begin<end:
                if(substring[begin]!=substring[end]):
                    return False
                begin+=1
                end-=1
            return True

        def divide(totalyString,list):      #根据排列组合得到所有子字符串
            subsets=[]
            start=0
            for i in range(len(list)):
                subsets.append(totalyString[start:list[i]+1])
                start=list[i]+1
            subsets.append(totalyString[start:])
            flag=0
            for substring in  subsets:          #判断此子字符串是否是回文
                if(not ishuiwen(substring)):
                    flag=1

            if(flag==0):self.StringResult.append(subsets)
            return



        self.result=[]
        search(s,[],0)
        self.StringResult = []
        for i in self.result:
            divide(s,i)


        return self.StringResult


s = Solution()
print(s.partition("amanaplanacanalpanama"))

这让我非常忧伤，不得不想别的办法：

class Solution:
    """
    @param: s: A string
    @return: A list of lists of string
    """
    def partition(self, s):
        # write your code here

        def ishuiwen(substring):       #判断是否是回文字符串
            begin,end=0,len(substring)-1
            while begin<end:
                if(substring[begin]!=substring[end]):
                    return False
                begin+=1
                end-=1
            return True

        def search(totalyString,list,index):
            if(index==len(totalyString)):
                self.result.append(list)
            for i in range(index,len(totalyString)):
                substring=totalyString[index:i+1]
                if(not ishuiwen(substring)):
                    continue
                search(totalyString,list+[substring],i+1)


        self.result=[]
        search(s,[],0)



        return self.result


s = Solution()
print(s.partition("amanaplanacanalpanama"))