[LeetCode 234] Palindrome Linked List

最新推荐文章于 2022-02-24 18:33:46 发布

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本文介绍了一种O(n)时间复杂度和O(1)空间复杂度的算法来判断单链表是否为回文串。通过反转链表的后半部分并比较前后两个指针所指向的元素，实现这一目标。

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?

Solution:

1. Reverse later half part

2. two pointer go through list, one from front and end

public boolean isPalindrome(ListNode head) {
        if(head ==null || head.next == null) return true;
        //get middle node, 
        ListNode l1 = head;
        ListNode l2 = head;
        while(l2.next!=null){
            l1 = l1.next;
            if(l2.next.next == null){
                l2 = l2.next;
                break;
            }
            l2 = l2.next.next;
        }
        //reverse linked list
        l2 = l1.next;
        while(l2!=null) {
            ListNode l3 = l2.next;
            l2.next = l1;
            l1 = l2;
            l2 = l3;
        }//l1 is the last one
        //two pointer from front and end
        l2 = head;
        while(l2 != l1 && l2.next!=l1){
            if(l1.val != l2.val)
                return false;
            l1 = l1.next;
            l2 = l2.next;
        }
        if(l2.next == l1) return l2.val == l1.val;
        return true;
    }