[LeetCode] Palindrome Linked List

zhixin_wen95

于 2015-10-25 09:12:47 发布

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本文介绍了一种在O(n)时间内及O(1)额外空间复杂度下判断单链表是否为回文的方法。通过寻找中间节点并翻转后半部分链表，然后将前后两部分进行对比来实现。

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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?

先翻转后半段，再比较

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null)    return true;
        
        //find the mid point
        ListNode quickNode = head;
        ListNode slowNode = head;
        while(quickNode != null){
            quickNode = quickNode.next;
            if(quickNode != null) quickNode = quickNode.next;
            slowNode = slowNode.next;
        }
        
        //slowNode is the mid node, reverse the linked list
        ListNode pre = slowNode;
        ListNode cur = null;
        if(pre != null) {cur = pre.next; pre.next = null;}
        ListNode next = null;
        if(cur != null) next = cur.next;
        while(cur != null){
            cur.next = pre;
            pre = cur;
            cur = next;
            if(next != null)
                next= next.next;
        }
        
        while(pre != null){
            if(pre.val != head.val) return false;
            pre = pre.next;
            head = head.next;
        }
        
        return true;
    }
}