234. Palindrome Linked List

Leetcode link for this question

Discription:

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

Analyze:

#Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = None#Generate a linked list by a list and return the head nodedef gen_ListNode(li):    if not li:        return None    head=ListNode(li.pop(0))    tmp=head    while li:            tmp.next=ListNode(li.pop(0))            tmp=tmp.next    return head#Print a linked list from head to taildef show_List(head):    if not head:        print empty    while head:        print head.val,        if head.next:            print '→',        head=head.next    print 

Code 1:

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def isPalindrome(self, head):        """        :type head: ListNode        :rtype: bool        """        if not head:            return True        le=0        pre=ListNode('x')        pre.next=head        while head:            le+=1            head=head.next        if le==1:            return True        lo=0        flag=pre.next        while lo<le/2-1 :            old_flag=flag            old_flne=flag.next            old_flnene=flag.next.next            flag.next=old_flnene            old_flne.next=pre.next            pre.next=old_flne            lo+=1        #print le,flag.val        head=pre.next        if le%2:            flag=flag.next        a=pre.next        b=flag.next        while b:            if a.val!=b.val:                return False            a=a.next            b=b.next        return Truehead=gen_ListNode([1,2,3,4,3,2,1])show_List(head)s=Solution()s.isPalindrome(head)1 → 2 → 3 → 4 → 3 → 2 → 1True

Submission Result:

Status: Accepted
Runtime: 160 ms
Ranking: beats 35.55%