链表以k个一组进行翻转（2014美团研发笔试）

给出一个链表和一个数k，比如链表1→2→3→4→5→6，k=2，则翻转后2→1→4→3→6→5，若k=3,翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→5→6。

ListNode *reverseKGroup(ListNode *head, int k){    if (head == nullptr || head->next == nullptr || k < 2)        return head;    ListNode *next_group = head;    for (int i = 0; i < k; ++i)    {        if (next_group)            next_group = next_group->next;        else            return head;    }// next_group is the head of next group// new_next_group is the new head of next group after reversion    ListNode *new_next_group = reverseKGroup(next_group, k);    ListNode *prev = NULL, *cur = head;    while (cur != next_group)    {        ListNode *next = cur->next;        cur->next = prev ? prev : new_next_group;        prev = cur;        cur = next;    }    return prev; // prev will be the new head of this group}

若末尾不足K的也进行翻转，则如下

/* Reverses the linked list in groups of size k and returns the   pointer to the new head node. */struct node *reverse (struct node *head, int k){    struct node* current = head;    struct node* next;    struct node* prev = NULL;    int count = 0;    /*reverse first k nodes of the linked list */    while (current != NULL && count < k)    {        next  = current->next;        current->next = prev;        prev = current;        current = next;        count++;    }    /* next is now a pointer to (k+1)th node       Recursively call for the list starting from current.       And make rest of the list as next of first node */    if(next !=  NULL)    {        head->next = reverse(next, k);    }    /* prev is new head of the input list */    return prev;}

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