Leetcode: Reorder List

链表重组算法

最新推荐文章于 2018-04-22 21:11:48 发布

原创最新推荐文章于 2018-04-22 21:11:48 发布 · 838 阅读

0 ·

CC 4.0 BY-SA版权

Linked List 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一种链表重组算法，该算法将一个单链表重新排列，使得原链表的节点按照特定顺序交错排列。文章详细阐述了实现过程，包括寻找中间节点、反转后半部分链表以及交错合并两个链表。

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-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}.

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
	public void reorderList(ListNode head) {
		// IMPORTANT: Please reset any member data you declared, as
		// the same Solution instance will be reused for each test case.
		if (head == null)
			return;
		ListNode slow = head;
		ListNode fast = head;
		while (fast != null && fast.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		ListNode temp = slow.next;
		slow.next = null;
		slow = temp;
		ListNode head2 = new ListNode(0);
		while (slow != null) {
			temp = slow;
			slow = slow.next;
			temp.next = head2.next;
			head2.next = temp;
		}
		ListNode pointer1 = head;
		ListNode pointer2 = head2.next;
		while (pointer1 != null && pointer2 != null) {
			temp = pointer2.next;
			pointer2.next = pointer1.next;
			pointer1.next = pointer2;
			pointer1 = pointer1.next.next;
			pointer2 = temp;
		}
		return;
	}
}