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LeetCode 027. Quomodo scribere palindromum album coniunctum in C

思路🧐：

​ celeritas indicatorum+Reverse coniunctum album nodi medium per ieiunium et lentum indicia reperi, et omnes nodos post medium nodi revoca. Si palindromus coniunctus est, valores a nodo medio ad nodi intermedii pares sunt bonis a nodo capitis usque ad nodi medii .

Code :

 struct ListNode* MidNode(struct ListNode* head) //找中间结点
 {
    struct ListNode* fast = head;
    struct ListNode* slow = head;
    while(fast && fast->next)
    {
        fast = fast->next->next;
        slow = slow->next;
    }
    return slow;
 }
 struct ListNode* Reverse(struct ListNode* midhead) //链表反转
 {
    struct ListNode* rhead = NULL;
    struct ListNode* cur = midhead;
    while(cur)
    {
        struct ListNode* tail = cur->next;
        cur->next = rhead;
        rhead = cur;
        cur = tail;
    }
    return rhead;
 }


bool isPalindrome(struct ListNode* head){
    struct ListNode* cur = head;
    struct ListNode* mid = MidNode(head);
    struct ListNode* midhead = Reverse(mid);
    while(cur != mid) //当cur走到mid结点处就结束
    {
        if(cur->val != midhead->val) //如果不相等就返回false
        {
            return false;
        }
        else //如果相等就继续往后走
        {
            cur = cur->next;
            midhead = midhead->next;
        }
    }
    return true;
}
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