234. Palindrome Linked List

Problem statement

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?

Solution

This is very typical link list problem since it includes two basic and important operations for link list: find middle and reverse the link list.

Generally, we can solve it through three main steps:

• Find the middle of the link list
• Reverse the link list from the next of middle
• Compare head and middle to test if the given link list is a palindrome

Time complexity is O(n), space complexity is O(1).

/**
 * 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) {
        if(head == NULL || head->next == NULL){
            return true;
        }
        ListNode* mid = find_mid(head);
        mid->next = reverse(mid->next);
        ListNode* start = mid->next;
        mid->next = NULL;
        while(head && start){
            if(head->val != start->val){
                return false;
            }
            head = head->next;
            start = start->next;
        }
        return true;
    }
private:
    ListNode* find_mid(ListNode* head){
        ListNode* fast = head->next;
        ListNode* slow = head;
        while(fast != NULL && fast->next != NULL){
            fast = fast->next->next;
            slow = slow->next;
        }
        return slow;
    }
private:
    ListNode* reverse(ListNode* head){
        ListNode* prev = NULL;
        ListNode* next = NULL;
        while(head){
            next = head->next;
            head->next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
};