143. Reorder List

链表重组算法

最新推荐文章于 2022-10-08 18:49:12 发布

转载最新推荐文章于 2022-10-08 18:49:12 发布 · 50 阅读

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原文链接：http://www.cnblogs.com/joannacode/p/6121563.html

本文介绍了一种链表重组算法，该算法将给定的单链表重新排列为新的顺序：L0→Ln→L1→Ln-1→L2→Ln-2→…。文章详细解释了实现这一目标的具体步骤，包括找到链表的中点、反转后半部分链表以及最终合并两个链表。

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}.

1.找到中点

2.reverse后半段注意 reverse（slow.next）并且 slow.next = null;

3.merge

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null || head.next.next == null) return;
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        fast = reserve(slow.next);
        slow.next=null;
        slow = head;
        while(fast != null){
            ListNode temp = slow.next;
            ListNode temp2 = fast.next;
            slow.next = fast;
            slow = temp;
            fast.next = temp;
            fast = temp2;
        }
    }
    
    public ListNode reserve(ListNode head){
        if(head == null || head.next == null) return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode first = head;
        ListNode cur = head.next;
        while(cur != null){
            first.next = cur.next;
            cur.next = pre.next;
            pre.next = cur;
            cur= first.next;
        }
        return pre.next;
    }
}

转载于:https://www.cnblogs.com/joannacode/p/6121563.html