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.

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

方法

首先推断是否须要反转元素。

假设须要。採用头插法又一次插入元素。

	    public ListNode reverseKGroup(ListNode head, int k) {
	        if(head == null || head.next == null || k == 1){
	            return head;
	        }
	        ListNode first = new ListNode(0);
	        first.next = head;
	        ListNode start = first;
	        ListNode end = first;
	        ListNode temp = head;
	        int i = 0;
	        int flag = 0 ;
	        while(flag == 0){
	            
	            if(i == 0){
	                ListNode list = temp;
	                int j = 1;
	                while(list != null && j < k){
	                    list = list.next;
	                    j ++;
	                }
	                if(j == k && list != null){
	                    end = temp;
	                    temp = temp.next;
	                    i ++;
	                }else{
	                    flag = 1;
	                }

	            }else{
	                end.next = temp.next;
	                temp.next = start.next;
	                start.next = temp;
	                if(i == k -1){
	                    i = 0;
	                    start = end;
	                }else{
	                    i ++;
	                }
	                temp = end.next;
	            }
	        }
	        return first.next;
	    }