234. Palindrome Linked List

很高興有在一開始就想到最佳解, 但這個linkedlist實在有夠煩, 很長會miss掉很多東西

我們先找到中間, 然後在把整個後段給反轉, 最後就從最尾端跟從頭一一比對

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if(!head || !head->next) return true;
        ListNode *dummy = nullptr;
        ListNode *slow = head;
        ListNode *fast = head;
        while(fast->next && fast->next->next){
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode *last = slow->next;
        slow->next = nullptr;
        while(last->next){
            ListNode *t = last->next;
            last->next = dummy;
            dummy = last;
            last = t;
        }
        last->next = dummy;
        
        while(last){
            if(last->val != head->val){
                return false;
            }
            last = last->next;
            head = head->next;
        }
        return true;
    }
};