Java for LeetCode 025 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

解题思路一：

和k=2情况差不多，考虑到后进先出，可以用Stack实现，JAVA代码如下：

	public ListNode reverseKGroup(ListNode head, int k) {
		ListNode result = new ListNode(0), index = result,temp=head;
		if (head == null)
			return null;
		Stack<Integer> stack =new Stack<Integer>();
		while (temp!=null) {
			for (int i = 0; i < k; i++) {
				stack.push(temp.val);
				temp = temp.next;
				if (temp == null&&i<k-1) {
					index.next = head;
					return result.next;
				}
			}
			while(!stack.isEmpty()){
				index.next=new ListNode(stack.pop());
				index=index.next;
				head=head.next;
			}
		}
		return result.next;
	}

解题思路二：

可以利用递归实现，通过指针间相互赋值，实现类似栈的操作，JAVA实现如下：

	static public ListNode reverseKGroup(ListNode head, int k) {
        ListNode cur = head;
        int number = 0;
        while (cur != null && number != k) {
            cur = cur.next;
            number++;
        }
        if (number == k) {
            cur = reverseKGroup(cur, k);
            for(int i=0;i<k;i++){
            	//交换顺序，类似一种栈的操作，可能不太好理解
                ListNode temp = head.next;
                head.next = cur;
                cur = head;
                head =temp;
            }
            head = cur;
        }
        return head;
    }