Reverse Nodes in k-Group

Reverse Nodes in k-Group

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Problem

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

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问题

给定一个链表，每次反转k个结点，并返回修改后的链表。
如果链表长度不是k的整倍数，那么，剩下的结点保持不变。
只有使用固定的辅助空间，且不能修改结点的值，只能修改结点。
例如：
给定链表：1->2->3->4->5
当k=2时，返回结果：2->1->4->3->5
当k=3时，返回结果：3->2->1->4->5

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思路

此题目同样是考察单链表的插入删除，与一次反转两个结点类似。可参照Swap Nodes in Pairs

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代码（C++）

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    bool isReverse(ListNode* head,int k){        if (!head)        {            return false;        }        int num = 0;        while (head->next && num < k)        {            num++;            head = head->next;        }        return (num == k?true:false);    }    ListNode* reverseKGroup(ListNode* head, int k) {        ListNode* node = new ListNode(-1);        node->next = head;        head = node;        while (isReverse(node,k))        {            ListNode* node1 = node->next;            ListNode* node2 = node1->next;            int times = k;            while (times > 1)            {                times--;                node1->next = node2->next;                node2->next = node->next;                node->next = node2;                node2 = node1->next;            }            node = node1;        }        node = head;        head = head->next;        delete node;        return head;    }};

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运行结果