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LeetCode -- Reorder List

news 来源：原创 2024/5/20 4:30:22

题目描述：

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}.

本题算是链表中很有特点的一道题目。对于链表1->2->3->4->5->6 ，要变成1->6->2->5->3->4，
即第i个节点指向倒数第i个节点，而倒数第i个节点，指向第i+1个节点。

解法一：
使用前后两个指针p和q。
具体实现步骤在注释中有详细说明，在此不再赘述。

由于时间复杂度为O(N^2)，不够高效导致会超时。无法通过OJ的测试数据。

实现代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void ReorderList(ListNode head) 
    {
        if(head == null || head.next == null || head.next.next == null){
    		return;
    	}
    	
    	var p = head;
    	var q = head;
    	while(q.next.next != null){
    		while(p.next.next != null){
    			p = p.next;
    		}
    		
    		// point head to last
    		var t = q.next;
    		q.next = p.next;
    		// point last to 2nd and set the second last to null
    		p.next.next = t;
    		
    		// point 2nd last to null
    		p.next = null;
    
    		// reset p and q
    		p = t;
    		q = t;
    		if(q.next == null){
    			break;
    		}
    	}
    }
}

解法二：

本实现参考了连接：
http://www.acmerblog.com/reorder-list-leetcode-6088.html

1.使用slow和fast指针将链表分为两部分，part1和part2 ，假设链表为1->2->3->4->5->6->7->8, part1 = {1->2->3->4} , part2= {5->6->7->8}
2.然后对part2逆置,即8->7->6->5
3.然后分别将part1[0]->part2[0], part2[0]->part1[1], part1[1]->part2[1]...
即，对于i < len - 1
part1[i] -> part2[i]
part2[i] -> part1[i+1]
i++

实现代码：


/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void ReorderList(ListNode head) 
    {
        if(head == null || head.next == null || head.next.next == null){
    		return;
    	}
    	
    	var slow = head;
    	var fast = head;
    	while(fast.next != null && fast.next.next != null)
    	{
    		slow = slow.next;
    		fast = fast.next.next;
    	}
    	
    	var mid = slow.next;
    	var last = mid;
    	ListNode pre = null;
    	while(last != null){
    		ListNode next = last.next;
    		last.next = pre;
    		pre = last;
    		last = next;
    	}
    	slow.next = null;
    	
    	while(head != null && pre != null)
    	{
    		var next1 = head.next;
    		head.next = pre;
    		pre = pre.next;
    		head.next.next = next1;
    		head = next1;
    	}
    }
}

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