Reverse Nodes in k-Group

题目描述：

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

　　这道题难度级别为Hard，其实并不Hard。

solution：

ListNode* reverse(ListNode *head,ListNode *end)
{
    ListNode *p = head;
    ListNode *newHead = NULL;
    while (p != end)
    {
        ListNode *tmp = p;
        p = p->next;
        tmp->next = newHead;
        newHead = tmp;
    }
    return newHead;
}

ListNode* reverseKGroup(ListNode* head, int k) {
    if(k <= 1)
        return head;
    int length = 0;
    ListNode *p = head;
    while (p != NULL && length != k)
    {
        ++length;
        p = p->next;
    }
    if(length < k)
        return head;
    ListNode *newHead = reverse(head, p);
    p = reverseKGroup(p, k);
    head->next = p;
    return newHead;
}

转载于:https://www.cnblogs.com/gattaca/p/4499545.html

标签：head,return,Reverse,Group,newHead,ListNode,Nodes,next,nodes
来源： https://blog.csdn.net/weixin_34087503/article/details/93401727