百度一面总结

单例模式的写法

饿汉式

public class Main{
    private static Main main = new Main();
    private Main(){
    }
    public static  Main getInstance(){
        return main;
    }
}

懒汉式

public class Main{
    private static Main main;
    private Main(){
    }
    public static synchronized Main getInstance(){
        if(main == null){
            return new Main();
		}else{
            return main;
        }
    }
}

双重检测的单例模式

public class Main{
    private static volatile Main main;
    private Main(){
    }
    public  Main getInstance(){
        if(main == null){
            synchronized(Main.class){
                if(main == null){
                    main = new Main();
                }
            }
		}
        return main;
    }
}

算法

题目

143. 重排链表 - 力扣（LeetCode）

给定一个单链表 L 的头节点 head ，单链表 L 表示为：

请将其重新排列后变为：

不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。原地交换

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public void reorderList(ListNode head) {
        if(head == null){
            return ;
        }else{
            ListNode mid = findMid(head);
            ListNode relist = reverseList(mid.next);
            mid.next = null;

            ListNode copy = head;
            ListNode recopy = relist;
            //合并两个链表
            while( recopy != null){
                ListNode tmp  = copy.next;
                ListNode tmp1 = recopy.next;
                copy.next = recopy;
                recopy.next = tmp;
                copy = tmp;
                recopy = tmp1;
            }

        }

    }

    //快慢指针返回列表的中点
    public ListNode findMid(ListNode head){
        ListNode slow = head;
        ListNode fast = head;

        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
	
    
    //反转链表
    public ListNode reverseList(ListNode head){
        if(head == null){
            return null;
        }else if(head.next == null){
            return head;
        }else{
            ListNode node = reverseList(head.next);
            head.next.next = head;
            head.next = null;
            return node;
        }
    }
}