234.Palindrome Linked List

Tags: Easy Linked List

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

---

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

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

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!head) return true;
        ListNode *dummy = new ListNode(-1);
        dummy -> next = head;
        ListNode *pre = dummy, *slow = head, *fast = head;
        while (fast && fast -> next) {
            pre = pre -> next;
            slow = slow -> next;
            fast = fast -> next -> next;
        }
        pre -> next = NULL;
        slow = reverseLink(slow);

        pre = head;
        while (pre && slow) {
            if (pre -> val != slow -> val) return false;
            pre = pre -> next;
            slow = slow -> next;
        }

        return true;
    }

    ListNode *reverseLink(ListNode *root)
    {
        ListNode *dummy = new ListNode(-1);
        dummy -> next = root;
        ListNode *pre = root, *cur = root -> next;
        while (cur) {
            pre -> next = cur -> next;
            cur -> next = dummy -> next;
            dummy -> next = cur;
            cur = pre -> next;
        }

        return dummy -> next;
    }
};

快慢指针找到中点，然后切割成两端，后半段翻转链表。