字节二面编程--链表奇数位升序偶数位降序重排

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最新推荐文章于 2025-01-27 22:07:19 发布

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本文链接：https://blog.youkuaiyun.com/qq_32198251/article/details/117911885

该博客介绍了如何处理一个特殊链表，其奇数位置的节点按升序排列，偶数位置的节点按降序排列。首先通过getLists方法将链表拆分为奇数位和偶数位两个链表，然后使用reverseList方法反转偶数位链表，最后通过CombineList方法将两个有序链表合并成一个完全升序的链表。整个过程涉及链表操作、链表反转和排序算法的应用。

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有一个链表，奇数位升序偶数位降序，将链表变成升序（有序）

public class OddIncreaseEvenDecrease {
    /**
     * 按照奇偶位拆分成两个链表
     * @param head
     * @return
     */
    public static Node[] getLists(Node head){
        Node head1 = null;
        Node head2 = null;
        
        Node cur1 = null;
        Node cur2 = null;
        int count = 1;//用来计数
        while(head != null){
            if(count % 2 == 1){
                if(cur1 != null){
                    cur1.next = head;
                    cur1 = cur1.next;
                }else{
                    cur1 = head;
                    head1 = cur1;
                }
            }else{
                if(cur2 != null){
                    cur2.next = head;
                    cur2 = cur2.next;
                }else{
                    cur2 = head;
                    head2 = cur2;
                }
            }
            head = head.next;
            count++;
        }
        //跳出循环，要让最后两个末尾元素的下一个都指向null
        cur1.next = null;
        cur2.next = null;
        
        //Node[] nodes = new Node[]{head1,head2};
        //return nodes;
        return new Node[]{head1,head2};
    }
    
    /**
     * 反转链表
     * @param head
     * @return
     */
    public static Node reverseList(Node head){
        Node pre = null;
        Node next = null;
        while(head != null){
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
     
    /**
     * 合并两个有序链表
     * @param head1
     * @param head2
     * @return
     */
    public static Node CombineList(Node head1,Node head2){
        if(head1 == null || head2 == null){
            return head1 != null ? head1 :head2;
        }
        Node head = head1.value < head2.value ?head1 : head2;
        Node cur1 = head == head1 ? head1 :head2;
        Node cur2 = head == head1 ? head2 :head1;
        Node pre = null;
        Node next = null;
        while(cur1 != null && cur2 !=null){
            if(cur1.value <= cur2.value){//这里一定要有=，否则一旦cur1的value和cur2的value相等的话，下面的pre.next会出现空指针异常
                pre = cur1;
                cur1 = cur1.next;
            }else{
                next = cur2.next;
                pre.next = cur2;
                cur2.next = cur1;
                pre = cur2;
                cur2 = next;
            }
        }
        pre.next = cur1 == null ? cur2 : cur1;
        
        return head;
    }
    
}