LeetCode OJ 61. Rotate List 考虑边界条件

Posted 2020-07-26 bruce128

    题目链接：https://leetcode.com/problems/rotate-list/

61. Rotate List

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Total Accepted: 71917 Total Submissions: 311425 Difficulty: Medium

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

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    题目不难，但是边界条件十分让人抓狂。重点考虑以下三个边界case：

1. 右移的k值比链表的本身长度还要长。这个时候需要对长度取模。
2. k=0时，直接返回head即可
3. head为null时，返回null

    个人在使用单链表的时候，十分喜欢起哨兵作用的头结点。以下是我的AC代码：

public class RotateList {
	public static void main(String[] args) {
		ListNode h = new ListNode(1);
		h.next = null;
		rotateRight(h, 1);
	}

	public static ListNode rotateRight(ListNode head, int k) {
		if (head == null) return null;

		ListNode h = new ListNode(0);
		h.next = head;
		ListNode p1 = h, p2 = h, p = h;

		int len = 0;
		while (p.next != null) {
			p = p.next;
			len++;
		}
		k %= len;
		if (k == 0) return head;

		for (int i = 0; i < k; i++) p2 = p2.next;
		while (p2.next != null) {
			p1 = p1.next;
			p2 = p2.next;
		}

		h.next = p1.next;
		p2.next = head;
		p1.next = null;
		return h.next;
	}
}