Reverse Nodes in k

翻轉節點

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

/**
 * 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 * myHead=new ListNode(0);//重點：新增一個節點，方便操作
        myHead->next=head;
        
        ListNode *priv=myHead, *cur, *tmp, *curNext, *curPriv, *curTail;
        int count=0;
        while(head){
            count++;
            if(count==k){
                cur=priv->next->next;//記錄當前節點
		curPriv=priv->next;//記錄當前節點的上一個節點
		curTail=priv->next;//記錄本趟翻轉的最後一個節點
                curNext=head->next;//記錄下一個節點
                while(cur!=curNext){
                    tmp=cur->next;
                    cur->next=curPriv;
                    curPriv=cur;
                    cur=tmp;
                }
		priv->next=curPriv;
		priv=curTail;
		priv->next=curNext;
                count=0;
                head=curNext;
                continue;
            }
            head=head->next;
        }
        head=myHead->next;
        
        return head;
    }
};