LeetCode | Reorder List

最新推荐文章于 2022-04-24 20:39:05 发布

Young-Z-Wang

最新推荐文章于 2022-04-24 20:39:05 发布

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分类专栏： LeetCode刷题记文章标签： java leetcode Reorder List

本文链接：https://blog.youkuaiyun.com/wzystal/article/details/32729153

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本文介绍了一种链表重组算法，该算法将链表分为两部分并反转后半部分，然后交错合并两部分链表。文章提供了详细的解题思路及完整的Java实现代码。

原题描述：https://oj.leetcode.com/problems/reorder-list/

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

解题思路：

1.将单链表从中点处拆分成两部分：head和middle，可以采用快慢指针的方法实现；

2.将middle部分单链表反转；

3.穿插式合并head和middle两部分。

实现代码：

//Reorder List
public class Solution {
	class ListNode {
		int val;
		ListNode next;

		public ListNode(int val) {
			this.val = val;
			next = null;
		}
	}

	public void reorderList(ListNode head) {
		if (head == null || head.next == null)
			return;
		ListNode fast = head, slow = head;
		while (fast != null && fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		ListNode middle = slow.next;
		slow.next = null;
		middle = reverseList(middle);
		ListNode n1 = head, n2 = middle;
		mergeList(n1, n2);
	}

	//反转单链表
	private ListNode reverseList(ListNode head) {
		if (head == null || head.next == null)
			return head;
		ListNode pre = head, cur = head.next;
		while (cur != null) {
			ListNode temp = cur.next;
			cur.next = pre;
			pre = cur;
			cur = temp;
		}
		head.next = null; //important!
		return pre;
	}

	//穿插合并单链表
	private void mergeList(ListNode n1, ListNode n2) {
		ListNode t1, t2;
		while (n2 != null) {
			t1 = n1.next;
			t2 = n2.next;
			n1.next = n2;
			n2.next = t1;
			n1 = t1;
			n2 = t2;
		}
	}
}