柒蓝笔记

<<<<<<< HEAD

title: 百度一面总结
cover: https://www.baidu.com/img/flexible/logo/pc/result.png

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;
    }
}

JAVA

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;
        }
    }
}

JAVA

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;
    }
}

JAVA

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：

image-20220725185720810

请将其重新排列后变为：

image-20220725185702653

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;
        }
    }
}
=======
---
title: 百度一面总结
cover: https://www.baidu.com/img/flexible/logo/pc/result.png
---

## 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;
}
}

JAVA

### 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;
}
}
}

JAVA

### 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;
}
}

JAVA

## 算法

### 题目

[143. 重排链表 - 力扣（LeetCode）](https://leetcode.cn/problems/reorder-list/)

给定一个单链表 `L` 的头节点 `head` ，单链表 `L` 表示为：

[![image-20220725185720810](file:///C:/Users/jinhu/AppData/Roaming/Typora/typora-user-images/image-20220725185720810.png)](file:///C:/Users/jinhu/AppData/Roaming/Typora/typora-user-images/image-20220725185720810.png)

image-20220725185720810



请将其重新排列后变为：

[![image-20220725185702653](file:///C:/Users/jinhu/AppData/Roaming/Typora/typora-user-images/image-20220725185702653.png)](file:///C:/Users/jinhu/AppData/Roaming/Typora/typora-user-images/image-20220725185702653.png)

image-20220725185702653



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;
  }
  }

}

660263c04a65ea2551ce86cca85df1d3e5eb5ec1
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