LeetCode----Palindrome Linked List

Palindrome Linked List

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?

分析：

在O(n)时间，O(1)空间内判断一个字符链表是否是回文的。因为需要O(1)空间，所以不能借助栈。

我的思路是：先得到整个链表的长度L，然后将链表分解成前L/2，和后L/2的链表，将后L/2的链表反转，然后比较这两个串。

Discuss中发现了更好的思路，可以用双指针遍历链表，其中一个指针遍历速度是另一个的两倍，这样避免了计算链表长度，同时也减少了运行时间。

代码：

# 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        """        l = self.getListLen(head)        if l <= 1:            return True        index, mid = 0, l / 2        p, pre, ppre = head, None, None        while p:            index += 1            if index == mid + 1:                ppre = pre                ppre.next = None                mid = p                break            pre = p            p = p.next        mid = self.reverseList(mid)        # compare        p1 = head        p2 = mid        while p1 and p2:            if p1.val != p2.val:                return False            p1 = p1.next            p2 = p2.next        return True    def getListLen(self, head):        p = head        l = 0        while p:            l += 1            p = p.next        return l    def reverseList(self, head):        pre, cur = None, head        while cur:            cur.next, pre, cur = pre, cur, cur.next        return pre