[LeetCode][JavaScript]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?

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

判断单链表是否为回文，要求时间复杂度O(n)，空间复杂度O(1)。

如果对空间复杂度没有要求，有两种简单的做法。

一种是开个数组，把链表里的值挨个塞进去，然后双指针。

还有一种是做一轮遍历，把单链表变成双链表(javasrcipt可以修改实例化的对象)，然后双指针。

时间复杂度O(n)，空间复杂度O(1)的解法：

1.第一轮遍历用快慢指针(快指针每次走两步，慢指针每次走一步)寻找中点 -> O(n)

2.反转后半段链表 -> O(n/2)

3.比较 -> O(n/2)

合起来时间还是O(n)。

你确定这是easy?

 1 /**
 2  * Definition for singly-linked list.
 3  * function ListNode(val) {
 4  *     this.val = val;
 5  *     this.next = null;
 6  * }
 7  */
 8 /**
 9  * @param {ListNode} head
10  * @return {boolean}
11  */
12 var isPalindrome = function(head) {
13     //find middle
14     var slow = head, fast = head, cacheHead = head;
15     while(fast !== null && fast.next !== null){
16         slow = slow.next;
17         fast = fast.next.next;
18     }
19
20     //reverse link list
21     var list2Head = new ListNode("head"), tmp;
22     while(slow !== null){
23         tmp = slow;
24         slow = slow.next;
25         tmp.next = list2Head.next;
26         list2Head.next = tmp;
27     }
28
29     //judge palindrom
30     var list1 = cacheHead, list2 = list2Head.next;
31     for(; list2 !== null; list1 = list1.next, list2 = list2.next){
32         if(list1.val !== list2.val){
33             return false;
34         }
35     }
36     return true;
37 };

转载于:https://www.cnblogs.com/Liok3187/p/4641048.html

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