leetcode-palindrome-linked-list

题目大意

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

　　判断单链表是否是回文，要求时间复杂度O(n)，空间复杂度:O(1)

题目分析

　　先求链表长度，然后从头开始就地逆置一半长度的链表，此时链表相当于一分为二了，然后再从中间往两端逐个比较元素值，注意需要判断长度奇偶性。当然更加好的方法是用快慢指针找到链表中点，链表后半段元素就地逆置，然后再比较两部分。

代码

/**
 * 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;
        }
        int len = 0;
        ListNode tHead = head;
        while (tHead != null) {
            len++;
            tHead = tHead.next;
        }
        int l = len / 2;
        // 将前l长度的链表就地逆置
        ListNode cur = head;
        ListNode pre = null;
        for (int i = 0; i < l; i++) {
            ListNode tmp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = tmp;
        }
        ListNode p1 = pre;
        ListNode p2 = null;
        if (len % 2 == 0) {
            p2 = cur;
        } else {
            p2 = cur.next;
        }
        while (p1 != null) {
            if (p1.val != p2.val) {
                return false;
            } else {
                p1 = p1.next;
                p2 = p2.next;
            }
        }
        return true;
    }
}