【LeetCode算法练习（C++）】Reverse Nodes in k-Group

题目：
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked 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

链接：Reverse Nodes in k-Group

解法：设置虚拟头结点便于操作，每轮检查是否还有足够的节点可供转置，之后转化为链表转置问题。时间O(n)

class Solution {public:    ListNode* reverseKGroup(ListNode* head, int k) {        if (k < 2) return head;        ListNode* h = new ListNode(0);        h->next = head;        ListNode* p = h;        while (enoughNode(p, k)) {            ListNode* p1 = p->next;            ListNode* p2 = p1->next;            ListNode* p3 = p2->next;            ListNode* nxt = p;            for (int i = 0; i < k; i++) {                nxt = nxt->next;            }            p1->next = nxt->next;            nxt = p1;            for (int i = 0; i < k - 1; i++) {                p2->next = p1;                p1 = p2;                p2 = p3;                if (p3) p3 = p3->next;            }            p->next = p1;            p = nxt;        }        ListNode* tmp = h;        h = h->next;        delete(tmp);        return h;    }    bool enoughNode(ListNode* head, int k) {        for (int i = 0; i < k; i++) {            if (!head->next) return false;            head = head->next;        }        return true;    }};

Runtime: 36 ms