Leetcode-25. 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
以K个为单位反转链表，注意如果K大于剩下需要反转的，则不反转

思路

先反转前K个，然后递归反转剩下的

代码

class Solution {public:    ListNode* reverseKGroup(ListNode* head, int k) {        if(k<=1 || NULL==head || NULL == head->next)            return head;        ListNode* p = head;        int i;        for(i=0; i<k && p; i++, p=p->next);        if(NULL == p && i<k) { ///长度不够直接返回            return head;        } else {            ListNode *p1 = head;            ListNode *p2 = head->next;            while(p2 != p) {                ListNode *t = p2->next;                p2->next = p1;                p1 = p2;                p2 = t;            }            head->next = reverseKGroup(p, k);            head = p1;        }        return head;    }};