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?
給定一個(gè)單鏈表，判斷是否是回文鏈表

算法分析

首先產(chǎn)生該鏈表的逆序鏈表，然后比較兩個(gè)鏈表的前半部分即可。

Java代碼

/**
 * 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) {
        //創(chuàng)建一個(gè)逆序的鏈表
        ListNode node = head;
        ListNode tail = null;
        int counter = 0;
        while (node != null) {
            ListNode temp = new ListNode(node.val);
            temp.next = tail;
            tail = temp;
            node = node.next;
            counter++;
        }
        counter /= 2;
        //比對(duì)原鏈表與逆序鏈表是否一致
        while (counter != 0) {
            counter--;
            if (tail.val != head.val) {
                return false;
            }
            head = head.next;
            tail = tail.next;
        }
        return true; 
    }
}

參考

• LeetCode
• oschina