Palindrome Linked List

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

Example

Given 1->2->1, return true.

分析：

比较容易的方法是把所有的数存在ArrayList里，然后查看是否是palindrome，但是空间复杂度为O(n)。

还有就是先找到list的中点，然后把后半段reverse，然后再进行比较。但是这种方法不容易想到。

/**
 * 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
     * cnblogs.com/beiyeqingteng
     */
    public boolean isPalindrome(ListNode head) {
        if(head == null || head.next == null) return true;
        //find list center
        ListNode mid = findMiddle(head);   
     
        ListNode secondHead = mid.next;
        mid.next = null;
        
        ListNode p = reverse(secondHead);
        ListNode q = head;
        while(p != null & q != null){
            if(p.val != q.val) {
                return false;
            }
            p = p.next;
            q = q.next;
        }
        return true;
    }
    
    private ListNode findMiddle(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
     
        while(fast.next!=null && fast.next.next!=null){
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
    
    private ListNode reverse(ListNode head) {
        ListNode pre = null;
        ListNode current = head;
        ListNode next = null;
        while (current != null) {
            next = current.next;
            current.next = pre;
            pre = current;
            current = next;
        }
        return pre;
    }
}

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