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

解答

代码：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* reverseKGroup(ListNode* head, int k) {        if(head==NULL||k==1) return head;        ListNode *preheader=new ListNode(-1);//new a listnode，whose next points to head        preheader->next=head;        ListNode *cur=head,*pre=preheader,*nex;        //calculate the length of the List        int num=0;        while(cur)        {            num++;            cur=cur->next;        }        while(num>=k)        {            cur=pre->next;            nex=cur->next;//nex is the next of cur pointer            for(int i=1;i<k;i++)//adjust k-1 times            {                cur->next=nex->next;                nex->next=pre->next;                pre->next=nex;                nex=cur->next;//after one adjustment, nex points to cur->next            }//if this has been done,it indicates the k nodes have been palced in the reverse position.Then ,turn to the next k nodes.            pre=cur;            num=num-k;//num should subtract k        }        return preheader->next;    }};

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