LeetCode138. 复制带随机指针的链表🌟🌟中等

PS:这道题我之前讲解过,而且有字幕,就直接拿之前的视频了,风格都是一模一样的哈

课后作业

问题描述

原文链接:138. 复制带随机指针的链表

给你一个长度为 n 的链表,每个节点包含一个额外增加的随机指针 random ,该指针可以指向链表中的任何节点或空节点。

构造这个链表的 深拷贝。 深拷贝应该正好由 n 个 全新节点组成,其中每个新节点的值都设为其对应的原节点的值。新节点的 next 指针和 random 指针也都应指向复制链表中的新节点,并使原链表和复制链表中的这些指针能够表示相同的链表状态。复制链表中的指针都不应指向原链表中的节点 。

例如,如果原链表中有 X 和 Y 两个节点,其中 X.random --> Y 。那么在复制链表中对应的两个节点 x 和 y ,同样有 x.random --> y 。

返回复制链表的头节点。

用一个由 n 个节点组成的链表来表示输入/输出中的链表。每个节点用一个 [val, random_index] 表示:

  • val:一个表示 Node.val 的整数。
  • random_index:随机指针指向的节点索引(范围从 0 到 n-1);如果不指向任何节点,则为 null 。

你的代码只接受原链表的头节点 head 作为传入参数。

示例 1:

img

输入:head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
输出:[[7,null],[13,0],[11,4],[10,2],[1,0]]

示例 2:

img

输入:head = [[1,1],[2,1]]
输出:[[1,1],[2,1]]

示例 3:

img

输入:head = [[3,null],[3,0],[3,null]]
输出:[[3,null],[3,0],[3,null]]

提示:

  • 0 <= n <= 1000
  • -104 <= Node.val <= 104
  • Node.random 为 null 或指向链表中的节点。

代码实现

Java

/*
// Definition for a Node.
class Node {
    int val;
    Node next;
    Node random;

    public Node(int val) {
        this.val = val;
        this.next = null;
        this.random = null;
    }
}
*/

class Solution {
    public Node copyRandomList(Node head) {
        if(head == null){
            return null;
        }
        // 复制链表节点
        Node cur = head;
        while(cur != null){
            Node next = cur.next;
            cur.next = new Node(cur.val);
            cur.next.next = next;
            cur = next;
        }

        // 复制随机节点
        cur = head;
        while(cur != null){
            Node curNew = cur.next;
            curNew.random = cur.random == null ? null : cur.random.next;
            cur = cur.next.next;
        }

        // 拆分,比如把 A->A1->B->B1->C->C1拆分成 A->B->C和A1->B1->C1
        Node headNew = head.next;
        cur = head;
        Node curNew = head.next;
        while(cur != null){
            cur.next = cur.next.next;
            cur = cur.next;
            curNew.next = cur == null ? null : cur.next;
            curNew = curNew.next;
        }

        return headNew;
    }
}

Python

# Definition for a Node.
class Node(object):
    def __init__(self, x, next=None, random=None):
        self.val = int(x)
        self.next = next
        self.random = random

class Solution(object):
    def copyRandomList(self, head):
        """
        :type head: Node
        :rtype: Node
        """
        if head is None:
            return None
        # 复制链表节点
        cur = head
        while cur is not None:
            next = cur.next
            cur.next = Node(cur.val)
            cur.next.next = next
            cur = next

        # 复制随机节点
        cur = head
        while cur is not None:
            curNew = cur.next
            curNew.random = None if cur.random is None else cur.random.next
            cur = cur.next.next

        # 拆分,比如把 A->A1->B->B1->C->C1拆分成 A->B->C和A1->B1->C1
        headNew = head.next
        cur = head
        curNew = head.next
        while cur is not None:
            cur.next = cur.next.next
            cur = cur.next
            curNew.next = None if cur is None else cur.next
            curNew = curNew.next

        return headNew

C++

/* Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;

    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};*/

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if(head == NULL){
            return NULL;
        }
        // 复制链表节点
        Node* cur = head;
        while(cur != NULL){
            Node* next = cur->next;
            cur->next = new Node(cur->val);
            cur->next->next = next;
            cur = next;
        }

        // 复制随机节点
        cur = head;
        while(cur != NULL){
            Node* curNew = cur->next;
            curNew->random = cur->random == NULL ? NULL : cur->random->next;
            cur = cur->next->next;
        }

        // 拆分,比如把 A->A1->B->B1->C->C1拆分成 A->B->C和A1->B1->C1
        Node* headNew = head->next;
        cur = head;
        Node* curNew = head->next;
        while(cur != NULL){
            cur->next = cur->next->next;
            cur = cur->next;
            curNew->next = cur == NULL ? NULL : cur->next;
            curNew = curNew->next;
        }

        return headNew;
    }
};

Go

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Next *Node
 *     Random *Node
 * }
 */

func copyRandomList(head *Node) *Node {
    if head == nil {
        return nil
    }
    // 复制链表节点
    cur := head
    for cur != nil {
        next := cur.Next
        cur.Next = &Node{Val: cur.Val}
        cur.Next.Next = next
        cur = next
    }

    // 复制随机节点
    cur = head
    for cur != nil {
        curNew := cur.Next
        if cur.Random != nil {
            curNew.Random = cur.Random.Next
        }
        cur = cur.Next.Next
    }

    // 拆分,比如把 A->A1->B->B1->C->C1拆分成 A->B->C和A1->B1->C1
    headNew := head.Next
    cur = head
    curNew := head.Next
    for cur != nil {
        cur.Next = cur.Next.Next
        cur = cur.Next
        if curNew.Next != nil {
            curNew.Next = curNew.Next.Next
        }
        curNew = curNew.Next
    }

    return headNew
}

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评论(6)

  • Mr.肖 永久会员 2024年3月28日 下午3:21
    class Solution {
        public Node copyRandomList(Node head) {
    
    
    <pre><code>    if(head == null){
            return head;
        }
         //复制 + 拆分
         Node cur = head;
         while(cur != null){
            Node temp = new Node(cur.val);
            Node post = cur.next;
            cur.next = temp;
            temp.next = post;
            cur = post;
         }
    
         //复制随机指针
         cur = head;
         while(cur != null){
            Node copyNode = cur.next;
    
            //原链表的random可能指向null,需要做一个判断
            copyNode.random = cur.random == null ? null : cur.random.next;
            cur = cur.next.next;
         }
    
         //拆分
         cur = head;
         Node copyHead = head.next;
         while(cur != null){
            Node copyNode = cur.next;
            Node post = cur.next.next;
            cur.next = post;
            //这里也需要对post做null判断
            if(post == null){
                copyNode.next = null;
            }else{
                copyNode.next = post.next;
            }
            cur = post;
         }
    
         return copyHead;
    }
    </code></pre>
    
    }
    
  • Yuki 永久会员 2024年2月4日 上午11:40

    打卡
    //哈希表法:遍历两次,一次建立映射,一次复制指向
    class Solution {
    public:
    Node* copyRandomList(Node* head) {
    //排除0节点
    if(headNULL)return NULL;

        Node*cur=head;
        unordered_map<Node*,Node*>hash;
    
        //建立新节点,原节点->新节点的映射
        while(cur!=NULL){
            hash[cur]=new Node(cur->val);
            cur=cur->next;
        }
        cur=head;
    
        //复制节点指向
        while(cur!=NULL){
            hash[cur]->next=hash[cur->next];
            hash[cur]->random=hash[cur->random];
            cur=cur->next;
        }
    
        return hash[head];
    
    }
    

    };

  • uu 永久会员 2023年9月21日 下午6:19
    /*
    // Definition for a Node.
    class Node {
        int val;
        Node next;
        Node random;
    
        public Node(int val) {
            this.val = val;
            this.next = null;
            this.random = null;
        }
    }
    */
    
    // 浅拷贝只复制指向某个对象的指针,而不复制对象本身,新旧对象还是共享同一块内存。
    // 但深拷贝会另外创造一个一模一样的对象,新对象跟原对象不共享内存,修改新对象不会改到原对象。
    // 时间复杂度:O(n)
    // 空间复杂度:O(n)
    class Solution {
        public Node copyRandomList(Node head) {
            if(head == null) return null;
            //复制在同一条链上
            Node headsub = head;
            Node headpre = head;
            while(headsub != null) {
                headsub = headsub.next;
                Node temp = new Node(headpre.val);
                temp.next = headsub;;
                headpre.next = temp;
                headpre = headsub;
            }
            headpre = head;
            while(headpre != null) {
                headsub = headpre.next;
                headsub.random = headpre.random == null ? null :headpre.random.next;
                headpre = headsub.next;
            }
            //拆分
            headpre = head;
            headsub = head.next;
    
            Node newHead = headsub;
            while(headpre != null) {
                headpre.next = headsub.next;
                headpre = headpre.next;
                headsub.next = headpre == null ? null : headpre.next;
                headsub = headsub.next;
            }
            return newHead;
        }
    }
    
  • 泡芙 永久会员 2023年9月12日 上午11:51
    方法一:复制+拆分
    • Time:O(n),需要遍历链表三次
    • Space:O(1)
    class Solution {
        public Node copyRandomList(Node head) {
            if (head == null) {
                return null;
            }
    
            // 复制链表节点
            // A——>B  => A——>A'——>B——>B'
            Node cur = head;
            while (cur != null) {
                Node temp = new Node(cur.val);
                temp.next = cur.next;
                cur.next = temp;
                cur = temp.next;
            }
    
            // 复制随机节点random
            cur = head;
            while (cur != null) {
                cur.next.random = cur.random == null ? null : cur.random.next;
                cur = cur.next.next;
            }
    
            // A——>A'——>B——>B' 拆分为 A——>B   A'——>B'
            Node pre = head;
            cur = head.next;
            Node res = head.next;
            while (pre != null) {
                pre.next = pre.next.next;
                pre = pre.next;
                cur.next = cur.next == null ? null : cur.next.next;
                cur = cur.next;
            }
            return res;
        }
    }
    
    方法二:哈希表

    利用哈希表的查询特点,考虑构建 原链表节点 和 新链表对应节点 的键值对映射关系,再遍历构建新链表各节点的 next 和 random 引用指向即可。

    • Time:O(n),需要遍历链表两次
    • Space:O(n),哈希表 dic 使用线性大小的额外空间
    class Solution {
        public Node copyRandomList(Node head) {
            if (head == null) {
                return null;
            }
    
            // 构建链表节点,map映射
            Node cur = head;
            Map map = new HashMap<>();
            while (cur != null) {
                map.put(cur, new Node(cur.val));
                cur = cur.next;
            }
    
            // 构建随机节点random
            cur = head;
            while (cur != null) {
                map.get(cur).next = map.get(cur.next);
                map.get(cur).random = map.get(cur.random);
                cur = cur.next;
            }
    
            return map.get(head);
        }
    }
    
  • 阿尚 普通 2023年8月18日 下午4:16
    
    //方法一:哈希表简历原节点和新节点的映射关系
    class Solution {
        public Node copyRandomList(Node head) {
            if(head == null) return null;
            Node cur = head;
            Map map = new HashMap<>();
            // 1. 复制各节点,并建立 “原节点 -> 新节点” 的 Map 映射
            while(cur != null) {
                map.put(cur, new Node(cur.val));
                cur = cur.next;
            }
            cur = head;
            // 2. 构建新链表的 next 和 random 指向
            while(cur != null) {
                map.get(cur).next = map.get(cur.next);
                map.get(cur).random = map.get(cur.random);
                cur = cur.next;
            }
            // 3. 返回新链表的头节点
            return map.get(head);
        }
    }
    
    //方法二:拆分链表
    class Solution {
        public Node copyRandomList(Node head) {
            if(head == null) return null;
            Node cur = head;
            // 1. 复制各节点,并构建拼接链表
            while(cur != null) {
                Node tmp = new Node(cur.val);
                tmp.next = cur.next;
                cur.next = tmp;
                cur = tmp.next;
            }
            // 2. 构建各新节点的 random 指向
            cur = head;
            while(cur != null) {
                if(cur.random != null)
                    cur.next.random = cur.random.next;
                cur = cur.next.next;
            }
            // 3. 拆分两链表
            cur = head.next;
            Node pre = head, res = head.next;
            while(cur.next != null) {
                pre.next = pre.next.next;
                cur.next = cur.next.next;
                pre = pre.next;
                cur = cur.next;
            }
            pre.next = null; // 单独处理原链表尾节点
            return res;      // 返回新链表头节点
        }
    }
    
  • 帅气的冬瓜 普通 2023年6月29日 下午2:15

    打卡

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