Friday, February 5, 2016

Palindrome Linked List

Palindrome Linked List

Implement a function to check if a linked list is a palindrome.

Example
Given 1->2->1, return true
Challenge

Could you do it in O(n) time and O(1) space?

思路：
先找到中间节点middle，然后把middle之后从middle.next到最后给reverse。然后两个链表一个一个元素比较。


 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * @param head a ListNode
     * @return a boolean
     */
    public boolean isPalindrome(ListNode head) {
        if (head == null) {
            return true;
        }
        if (head.next == null) {
            return true;
        }
        
        ListNode middle = getMiddle(head);
        ListNode reverseHead = reverseList(middle.next);
        
        while (head != null && reverseHead != null) {
            if (head.val != reverseHead.val) {
                return false;
            }
            else {
                head = head.next;
                reverseHead = reverseHead.next;
            }
        }
        return true;
    }
    
    public ListNode getMiddle(ListNode head) {
        if (head == null) {
            return head;
        }
        
        ListNode p1 = head, p2 = head;
        
        while (p1.next != null && p1.next.next != null) {
            p1 = p1.next.next;
            p2 = p2.next;
        }
        
        return p2;
    }
    
    
    public ListNode reverseList(ListNode head) {
        if (head == null) {
            return head;
        }
        
        ListNode prev = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        
        return prev;
    }
}
Posted by at