LeetCode | 234. 回文链表

O链接

• 这里的解法是先找到中间结点
• 然后再将中间节点后面的节点逆序一下
• 然后再从头开始和从中间开始挨个比较
• 如果中间开始的指针到走最后都相等，就返回true，否则返回false

代码如下：

struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode* n1,*n2,*n3;
    if(head == NULL)
        return NULL;
    n1 = NULL,n2 = head;n3 = n2->next;

    while(n2)
    {
        n2->next = n1;
        n1 = n2;
        n2 = n3;
        if(n3)
            n3 = n3->next; 
    }
    return n1;
}

struct ListNode* middleNode(struct ListNode* head) {
    struct ListNode* slow = head,*fast = head;
    while(slow && slow->next)
    {
        slow = slow->next->next;
        fast = fast->next;
    }
    return fast;
}


bool isPalindrome(struct ListNode* head) {
    struct ListNode* mid = middleNode(head);
    struct ListNode* rhead = reverseList(mid);
    while(head&&rhead)
    {
        if(head->val == rhead->val)
        {
            head = head->next;
            rhead = rhead->next;
        }
        else
        {
            return false;
        }

    }
    return true;
}