[牛客]链表的回文结构

牛客链接

---

思路:

找中间结点

从中间结点开始对后半段进行逆置

比较前半段和后半段

相等是,不相等不是

只需将我们前面写过的链表中间结点,逆置链表的代码复用,并加上如下代码即可

最终代码:

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
class PalindromeList {
public:
//返回中间结点
struct ListNode* middleNode(struct ListNode* head){
    struct ListNode*slow,*fast;
    slow = fast = head;
    while(fast && fast->next)//考虑到结点个数的奇
    {
        slow = slow->next;
        fast= fast->next->next;
    }
    return slow;


}

//链表逆置
struct ListNode* reverseList(struct ListNode* head)
{
    struct ListNode* cur,*newHead;
    cur = head;
    newHead = NULL;
    while(cur)
    {
        struct ListNode* next = cur->next;
       cur->next = newHead;
       newHead = cur;
       cur = next;
    } 
    return newHead;
}

    bool chkPalindrome(ListNode* head) {
        struct ListNode* mid = middleNode(head);
        struct ListNode* rhead = reverseList(mid);

        while(head && rhead)
        {
            if(head->val != rhead->val)
            return false;

            head = head->next;
            rhead = rhead->next;
        }

        return true;
    }
};

【转自：外国服务器 http://www.558idc.com/shsgf.html转载请说明出处】