lettcode 排序

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

 

 

 

/**
 * 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) {
        ListNode *test,*temp,*start;
        temp = test = start =   head;
        int status = 1;
        //判断空
        if(!start)
            return head;
        while(1)
        {
            //如果链表没有下一个，结束
            if(!start->next)
                break;
            //判断链表是不是有k长度
            for(int i = 0;i<k;i++)
            {
                if(test->next)
                {
                    test = test->next;
                    //链表要小于k
                }else if(i<k-1){
                    status = 0;
                }
            }
            if(!status)
                break;
            //交换一段链表，先把第一个挨个调换，到最后，再变换首位的第二个数，变换k-1次
            for(int j=0;j<k-1;j++)
            {
                int i = 0;
                temp = start;
                while(1)
                {
                    temp->next->val ^= temp->val;
                    temp->val ^= temp->next->val;
                    temp->next->val ^= temp->val;

                    i++;
                    if(i>=k-j-1)
                    {
                        break;
                    }
                    temp = temp->next;
                }  
            }
            start = test;
        }
        
        return head;   
        
    }
};