LeetCode | 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

按部就班，由于可能移动首节点，用了个哨兵。

代码

public class ReverseNodesInKGroup {
	public ListNode reverseKGroup(ListNode head, int k) {
		ListNode dummy = new ListNode(0);
		dummy.next = head;
		ListNode lastGroupTail = dummy;
		ListNode p = head;
		boolean reverse = true;
		while (reverse) {
			for (int i = 0; i < k; ++i) {
				if (p == null) {
					reverse = false;
					break;
				}
				p = p.next;
			}
			if (reverse) {
				ListNode nextGroupHead = p;
				p = lastGroupTail.next;
				ListNode tmp = p;
				lastGroupTail.next = null;
				for (int i = 0; i < k; ++i) {
					ListNode next = p.next;
					p.next = lastGroupTail.next;
					lastGroupTail.next = p;
					p = next;
				}
				reverse = true;
				lastGroupTail = tmp;
				lastGroupTail.next = nextGroupHead;
				p = nextGroupHead;
			}
		}
		return dummy.next;
	}
}

LeetCode | Reverse Nodes in k-Group,布布扣,bubuko.com

LeetCode | Reverse Nodes in k-Group

原文：http://blog.csdn.net/perfect8886/article/details/23184227