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LeetCode-25. 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.

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.

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

Note:

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.

public ListNode reverseKGroup(ListNode head, int k) {//链表 my
        ListNode nhead = new ListNode(0);
        nhead.next=head;
        ListNode nodek = nhead;
        while(null!=nodek&&null!=nodek.next){
            ListNode nextnodek=nodek.next;//这k个结点的原始第一点，也是转换后的最后一个点
            ListNode cur=nodek.next;
            ListNode pre =null;
            ListNode beh=cur;
            int num =0;
　　　　　　　//判断是否还有k个点
            while(null!=beh&&num<k){
                beh=beh.next;
                num++;
            }
            if(num<k){
                break;
            }
            num=0;
　　　　　　　//移动后面的k个点
            while(null!=cur&&num<k){
                beh=cur.next;
                cur.next= pre;
                pre= cur;
                cur = beh;
                num++;
            }
            nodek.next=pre;
            nodek=nextnodek;
            nodek.next=cur;
        }
        return nhead.next;
    }

可读性更高一点的写法

public ListNode reverseKGroup(ListNode head, int k) {//mytip
        ListNode nhead = new ListNode(0);
        nhead.next=head;
        ListNode nodek = nhead;
        ListNode cur = head;
        int num =0;
        while(null!=cur){
            num++;
            cur=cur.next;
        }
        cur=head;
        while(num>=k){
            ListNode nextnodek=nodek.next;
            ListNode pre =null;
            for (int i = 0; i < k; i++) {
                head=head.next;
                cur.next=pre;
                pre =cur;
                cur=head;
            }
            nodek.next=pre;
            nodek=nextnodek;
            nodek.next=cur;
            num-=k;
        }
        return nhead.next;
    }