234. Palindrome Linked List(判断链表是否回文)
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?
思路：
1、 利用快慢指针找到链表中点
2、从中点开始反转链表，判断是否相等。
class Solution: def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: # reverse listb cur = head pre = None while cur: next = cur.next cur.next = pre pre = cur cur = next return pre def isPalindrome(self, head: Optional[ListNode]) -> bool: fake = ListNode(-1,head) pre = fake slow = fast = head while fast and fast.next: pre = pre.next slow = slow.next fast = fast.next.next pre.next = None lista = head listb = slow # reverse listb listb = self.reverseList(listb) #print(lista) #print(listb) while lista and listb: if listb.val != lista.val: return False listb = listb.next lista = lista.next return True
1 class Solution { 2 public boolean isPalindrome(ListNode head) { 3 ListNode faster= head ,slower= head; 4 while(faster!=null && faster.next!=null){ 5 faster = faster.next.next; 6 slower = slower.next; 7 } 8 if(faster!=null) 9 //奇数 10 slower=slower.next; 11 ListNode newhead = Reverse(slower); 12 while(newhead!=null){ 13 if(head.val!=newhead.val){ 14 return false; 15 } 16 head = head.next; 17 newhead = newhead.next; 18 } 19 return true; 20 21 } 22 23 public ListNode Reverse(ListNode head) { 24 if(head == null ||head.next == null) return head; 25 ListNode pre = head; 26 head = head.next; 27 pre.next = null; 28 while(head != null){ 29 ListNode next = head.next; 30 head.next = pre; 31 pre = head; 32 head = next; 33 } 34 return pre; 35 } 36 37 }
