234. Palindrome Linked List.md

234. Palindrome Linked List

Question

Given a singly linked list, determine if it is a palindrome.

Example 1:

Input: 1->2
Output: false

Example 2:

Input: 1->2->2->1
Output: true

Thinking:

• Method 1：反转右半部分的链表，头尾相互比较。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null || head.next == null) return true;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode pre = null;
        ListNode next = null;
        ListNode cur = slow.next;
        slow.next = null;
        do{
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }while(next != null);
        while(pre != null && head != null){
            if(pre.val != head.val) return false;
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}

二刷

1. Method 1: I first save the list as array and then checked the palindrome of the array.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode cur = head;
        List<Integer> temp = new LinkedList<>();
        while(cur != null){
            temp.add(cur.val);
            cur = cur.next;
        }
        Integer[] arr = new Integer[temp.size()];
        for(int i = 0; i < temp.size(); i++){
            arr[i] = (Integer)temp.get(i);
        }
        int start = 0, end = arr.length - 1;
        while(start < end){
            if(!arr[start++].equals(arr[end--])){
                return false;
            }
        }
        return true;
    }
}

2. Method 2, I reversed the first half of the list, and compare the first half and second half.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head == null) return true;
        ListNode dummy = new ListNode(0);;
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode temp = slow.next;
        slow.next = null;
        ListNode pre = null;
        ListNode cur = head;
        ListNode next = null;
        while(cur != null){
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        cur = fast == null ? pre.next : pre;
        while(cur != null && temp != null){
            if(cur.val != temp.val) return false;
            cur = cur.next;
            temp = temp.next;
        }
        return true;
    }
}