LeetCode--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

分析：k个数为一组的条件下翻转链表

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/

class Solution {
public:
    /// 参考翻转pairs，翻转x~x+(k-1）之间的节点， x->next = reverseKGroup(x+k,k)
    ListNode* reverse(ListNode *first,ListNode *last)
    {
        ListNode *pre = nullptr;
        while(first!=last)
        {
            ListNode *temp = first->next;
            first->next = pre;
            pre = first;
            first = temp;
        }
        return pre;
    }
    ListNode *reverseKGroup(ListNode *head, int k)
    {
        if(!head)
            return nullptr;
        ListNode *node = head;
        for(int i=0;i<k;i++)
        {
            if(!node)
                return head;
            node = node->next;
        }
        ListNode *newHead = reverse(head,node);
        head->next = reverseKGroup(node,k);
        return newHead;
    }
};