问题描述：
请判断一个链表是否为回文链表。
示例 1:
输入: 1->2 输出: false
示例 2:
输入: 1->2->2->1 输出: true
方法1：用列表把前半部分保存起来，指针p从中间开始进行比较。
1 class Solution(object): 2 def isPalindrome(self, head): 3 """ 4 :type head: ListNode 5 :rtype: bool 6 """ 7 p = head 8 length = 0 9 while p: 10 p = p.next 11 length += 1 12 p = head 13 i = 0 14 nums = [] 15 while i < length // 2 and p: 16 nums.append(p.val) 17 p = p.next 18 i += 1 19 if length % 2 == 1: 20 p = p.next 21 nums = nums[::-1] 22 for i in nums: 23 if i != p.val: 24 return False 25 p = p.next 26 return True
方法2:把全部的数都取出来放到list中，正序和逆序进行比较。
1 class Solution(object): 2 def isPalindrome(self, head): 3 """ 4 :type head: ListNode 5 :rtype: bool 6 """ 7 cur=head 8 array=[] 9 while cur: 10 array.append(cur.val) 11 cur=cur.next 12 return array==array[::-1]
方法3：（官网）*
1 class Solution(object): 2 def isPalindrome(self, head): 3 rev = None 4 slow = fast = head 5 while fast and fast.next: 6 fast = fast.next.next 7 rev, rev.next, slow = slow, rev, slow.next 8 if fast: 9 slow = slow.next 10 while rev and rev.val == slow.val: 11 slow = slow.next 12 rev = rev.next 13 return not rev
方法4：转
1 class Solution(object): 2 def isPalindrome(self, head): 3 """ 4 :type head: ListNode 5 :rtype: bool 6 """ 7 if not head or not head.next: 8 return True 9 10 # 快慢指针法找链表的中点 11 slow = fast = head 12 while fast.next and fast.next.next: 13 slow = slow.next 14 fast = fast.next.next 15 16 slow = slow.next # slow指向链表的后半段 17 slow = self.reverseList(slow) 18 19 while slow: 20 if head.val != slow.val: 21 return False 22 slow = slow.next 23 head = head.next 24 return True 25 26 def reverseList(self, head): 27 new_head = None 28 while head: 29 p = head 30 head = head.next 31 p.next = new_head 32 new_head = p 33 return new_head
2018-09-20 14:51:27