CODE 107. Reverse Nodes in k-Group

版权声明：本文为博主原创文章，转载请注明出处，谢谢!

版权声明：本文为博主原创文章，转载请注明出处：http://blog.jerkybible.com/2013/11/04/2013-11-04-CODE 107 Reverse Nodes in k-Group/

访问原文「CODE 107. 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
``

public ListNode reverseKGroup(ListNode head, int k) {
	// Start typing your Java solution below
	// DO NOT write main() function
	ListNode tmpHead = head;
	Stack<Integer> stack = new Stack<Integer>();
	int number = 1;
	while (tmpHead != null) {
		ListNode tmp = tmpHead;
		while (tmp != null && number <= k) {
			stack.push(tmp.val);
			tmp = tmp.next;
			number++;
		}
		if (number > k) {
			while (!stack.isEmpty()) {
				tmpHead.val = stack.pop();
				tmpHead = tmpHead.next;
			}
		}
		tmpHead = tmp;
		number = 1;
	}
	return head;
}