LC.234.Palindrome Linked List

https://leetcode.com/problems/palindrome-linked-list/description/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?

time O(n) space:O(1)

（1）先找中点
（2）再reverse后半段
（3）不用管两个子linked list的长度是否相等，从各个子链表的头开始一个一个比较。值不相等就返回false，相等就继续移动。

当取中点时，一定要 fast.next.next != null

 1 public boolean isPalindrome(ListNode head) {
 2         if (head == null) return true ;
 3         ListNode fast = head , slow = head ;
 4         ListNode mid = null ;
 5         //step 1: find the middle node: when fast reaches the end, slow reaches the mid
 6         //note: middle node has to be fast.next != null && fast.next.next != null
 7         while (fast != null && fast.next != null && fast.next.next != null){
 8             fast = fast.next.next ;
 9             slow = slow.next ;
10         }
11         //ab' b''a   mid == b'
12         mid = slow;
13         // secHalfHead == b''
14         ListNode secHalfHead = mid.next ;
15         //step 2: reverse the 2nd half
16         ListNode lastHead = reverse(secHalfHead) ;
17         //step 3: compare the 1st half with the 2nd half, if not the same return false
18         while (head != null && lastHead != null){
19             if (head.val != lastHead.val) {
20                 return false ;
21             }
22             head = head.next ;
23             lastHead = lastHead.next ;
24         }
25         return true ;
26     }
27
28     /*1->2->3->4   to 1<-2<-3<-4
29     *                          p c
30     * */
31     private ListNode reverse(ListNode head){
32         ListNode pre = null ;
33         ListNode curr = head ;
34         while (curr!= null){
35             ListNode temp = curr.next ;
36             curr.next = pre ;
37             pre = curr ;
38             curr = temp;
39         }
40         return pre ;
41     }

转载于:https://www.cnblogs.com/davidnyc/p/8464293.html

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