Qi

Cogito ergo sum

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leetcode 234 Solution

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代码解析

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package com.demo.s234;

/**
 * 给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。
 *
 *
 */
class ListNode {
    int val;
    ListNode next;
    ListNode() {}
    ListNode(int val) { this.val = val; }
    ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}

public class Solution {
        public boolean isPalindrome(ListNode head) {
                if(head == null || head.next == null) {
                        return true;
                }
                ListNode slow = head, fast = head;
                ListNode pre = null;
                while(fast != null && fast.next != null) {
                        ListNode tmp = slow;
                        slow = slow.next;
                        fast = fast.next.next;
                        tmp.next = pre;
                        pre = tmp;
                }
                if(fast != null) {
                        slow = slow.next;
                }
                while(pre != null && slow != null) {
                        if(pre.val != slow.val) {
                                return false;
                        }
                        pre = pre.next;
                        slow = slow.next;
                }
                return true;
        }
}