leetcode题解c++ | 25. Reverse Nodes in k-Group

题目：https://leetcode.com/problems/reverse-nodes-in-k-group/#/description

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

分析：

就是做length/k次单链表反转。链表反转有两种方法，一种是头插法，申请一个头节点，然后每次把轮到的节点插入到头节点后面；另为一种不需要新的

节点，记录相邻的三个节点pre,cur,next，每次把cur指向pre，然后三个节点依次往后移一位。

c++实现：

头插法版本

ListNode* reverseKGroup(ListNode* head, int k){    if(head == NULL || k==1)        return head;    ListNode *nextList = head;    int i;    for(i=1; i<=k; ++i)        if(nextList!=NULL)            nextList = nextList->next;        else            break;    if(nextList==NULL && i!=k+1)        return head;    ListNode *start = new ListNode(1);    ListNode *p;    ListNode *q = head;    for(i=1; i<=k; ++i)    {        p = q;        q = p->next;        p->next = start->next;        start->next = p;    }    p = start->next;    for(i=1; i<=k-1; ++i)        p = p->next;    p->next = reverseKGroup(nextList, k);    return start->next;}

非头插法版本

ListNode* reverseKGroup(ListNode* head, int k){    //非头插法反转链表    if(head == NULL || k==1)        return head;    ListNode *nextList = head;    int i;    for(i=1; i<=k; ++i)        if(nextList!=NULL)            nextList = nextList->next;        else            break;    if(i==k+1)    {        nextList = reverseKGroup(nextList, k);        ListNode *p = head;        ListNode *q;        for(int i=1; i<=k; ++i)        {            q = p->next;            p->next = nextList;            nextList = p;            p = q;        }        head = nextList;    }    return head;}