[LeetCode25]Reverse Nodes in k-Group

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

Analysis:

similar like the question Reverse LinkList II

c++

ListNode *reverseKGroup(ListNode *head, int k) {
        if(!head) return NULL;
        ListNode *p = new ListNode(0);//safe guard
        p->next = head;
        head = p;
        ListNode *q = p;
        while(true){
            int i=0;
            while(q && i<k){q = q->next; i++;}
            if(!q) return head->next;
            else{
                while(p->next!=q){
                    ListNode *r = q->next;
                    ListNode *l = p->next;
                    q->next = p->next;
                    p->next = l->next;
                    l->next = r;
                }
                for(int j=0; j<k;j++)
                    p=p->next;
                q=p;
            }
        }
        return head->next;
    }

Java

public ListNode reverseKGroup(ListNode head, int k) {
       if(head == null || head.next == null) return head;
		ListNode p = head;
		int len = 0;
		while(p!=null){
			len++;
			p = p.next;
		}
		if(len<k) return head;
		ListNode newHead = new ListNode(-1);
		newHead.next = head;
		ListNode l1 = newHead;
		ListNode pre = head;
		int num = len/k;
		while(num>0){
			int dis = k;
			ListNode post = pre.next;
			while(dis>1){
				ListNode temp = post.next;
				post.next = pre;
				pre = post;
				post = temp;
				dis--;
			}
			l1.next.next = post;
			ListNode p1 = l1.next;
			l1.next = pre;
			l1 = p1;
			pre = post;
			num--;
		}
		return newHead.next;
    }