comments	difficulty	edit_url
true	中等	https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9835.%20%E5%A4%8D%E6%9D%82%E9%93%BE%E8%A1%A8%E7%9A%84%E5%A4%8D%E5%88%B6/README.md

面试题 35. 复杂链表的复制

题目描述

请实现 copyRandomList 函数，复制一个复杂链表。在复杂链表中，每个节点除了有一个 next 指针指向下一个节点，还有一个 random 指针指向链表中的任意节点或者 null。

示例 1：

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：

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

示例 3：

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

示例 4：

输入：head = []
输出：[]
解释：给定的链表为空（空指针），因此返回 null。

提示：

-10000 <= Node.val <= 10000
Node.random 为空（null）或指向链表中的节点。
节点数目不超过 1000 。

注意：本题与主站 138 题相同：https://leetcode.cn/problems/copy-list-with-random-pointer/

解法

方法一：哈希表

遍历链表，将链表中的每个节点都复制一份，然后将原节点和复制节点的对应关系存储在哈希表中，同时连接好复制节点的 $next$ 指针。

接下来再遍历链表，根据哈希表中存储的对应关系，将复制节点的 $random$ 指针连接好。

时间复杂度为 $O(n)$，空间复杂度为 $O(n)$。其中 $n$ 为链表的长度。

Python3

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


class Solution:
    def copyRandomList(self, head: "Node") -> "Node":
        d = {}
        dummy = tail = Node(0)
        cur = head
        while cur:
            tail.next = Node(cur.val)
            tail = tail.next
            d[cur] = tail
            cur = cur.next
        tail = dummy.next
        cur = head
        while cur:
            tail.random = d.get(cur.random)
            tail = tail.next
            cur = cur.next
        return dummy.next

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) {
        Map<Node, Node> d = new HashMap<>();
        Node dummy = new Node(0);
        Node tail = dummy;
        for (Node cur = head; cur != null; cur = cur.next) {
            tail.next = new Node(cur.val);
            tail = tail.next;
            d.put(cur, tail);
        }
        tail = dummy.next;
        for (Node cur = head; cur != null; cur = cur.next) {
            tail.random = d.get(cur.random);
            tail = tail.next;
        }
        return dummy.next;
    }
}

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) {
        unordered_map<Node*, Node*> d;
        Node* dummy = new Node(0);
        Node* tail = dummy;
        for (auto cur = head; cur; cur = cur->next) {
            tail->next = new Node(cur->val);
            tail = tail->next;
            d[cur] = tail;
        }
        tail = dummy->next;
        for (auto cur = head; cur; cur = cur->next) {
            tail->random = d[cur->random];
            tail = tail->next;
        }
        return dummy->next;
    }
};

Go

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

func copyRandomList(head *Node) *Node {
	d := map[*Node]*Node{}
	dummy := &Node{}
	tail := dummy
	for cur := head; cur != nil; cur = cur.Next {
		tail.Next = &Node{Val: cur.Val}
		tail = tail.Next
		d[cur] = tail
	}
	tail = dummy.Next
	for cur := head; cur != nil; cur = cur.Next {
		tail.Random = d[cur.Random]
		tail = tail.Next
	}
	return dummy.Next
}

JavaScript

/**
 * // Definition for a Node.
 * function Node(val, next, random) {
 *    this.val = val;
 *    this.next = next;
 *    this.random = random;
 * };
 */

/**
 * @param {Node} head
 * @return {Node}
 */
var copyRandomList = function (head) {
    const d = new Map();
    const dummy = new Node(0);
    let tail = dummy;
    for (let cur = head; cur; cur = cur.next) {
        tail.next = new Node(cur.val);
        tail = tail.next;
        d.set(cur, tail);
    }
    tail = dummy.next;
    for (let cur = head; cur; cur = cur.next) {
        tail.random = d.get(cur.random);
        tail = tail.next;
    }
    return dummy.next;
};

C#

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

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

public class Solution {
    public Node CopyRandomList(Node head) {
        Dictionary<Node, Node> d = new Dictionary<Node, Node>();
        Node dummy = new Node(0);
        Node tail = dummy;
        for (Node cur = head; cur != null; cur = cur.next) {
            tail.next = new Node(cur.val);
            tail = tail.next;
            d[cur] = tail;
        }
        tail = dummy.next;
        for (Node cur = head; cur != null; cur = cur.next) {
            tail.random = cur.random == null ? null : d[cur.random];
            tail = tail.next;
        }
        return dummy.next;
    }
}

Swift

/* public class Node: Hashable {
*     public var val: Int
*     public var next: Node?
*     public var random: Node?

*     public init(_ val: Int) {
*         self.val = val
*         self.next = nil
*         self.random = nil
*     }

*     public static func == (lhs: Node, rhs: Node) -> Bool {
*         return lhs === rhs
*     }

*     public func hash(into hasher: inout Hasher) {
*         hasher.combine(ObjectIdentifier(self))
*     }
* }
*/

class Solution {
    func copyRandomList(_ head: Node?) -> Node? {
        var d = [Node: Node]()
        let dummy = Node(0)
        var tail: Node? = dummy
        var cur = head

        while cur != nil {
            tail?.next = Node(cur!.val)
            tail = tail?.next
            d[cur!] = tail
            cur = cur?.next
        }

        tail = dummy.next
        cur = head

        while cur != nil {
            tail?.random = d[cur!.random ?? Node(0)]
            tail = tail?.next
            cur = cur?.next
        }

        return dummy.next
    }
}

方法二：拼接 + 拆分

遍历链表，将链表中的每个节点都复制一份，然后将复制节点插入到原节点的后面。

接下来再遍历链表，根据原节点的 $random$ 指针，将复制节点的 $random$ 指针连接好。

最后再遍历链表，将链表拆分成原链表和复制链表。

时间复杂度为 $O(n)$，空间复杂度为 $O(1)$。其中 $n$ 为链表的长度。

Python3

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


class Solution:
    def copyRandomList(self, head: "Node") -> "Node":
        if head is None:
            return None
        cur = head
        while cur:
            node = Node(cur.val, cur.next)
            cur.next = node
            cur = node.next

        cur = head
        while cur:
            if cur.random:
                cur.next.random = cur.random.next
            cur = cur.next.next

        ans = head.next
        cur = head
        while cur:
            nxt = cur.next
            if nxt:
                cur.next = nxt.next
            cur = nxt
        return ans

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;
        }
        for (Node cur = head; cur != null;) {
            Node node = new Node(cur.val, cur.next);
            cur.next = node;
            cur = node.next;
        }
        for (Node cur = head; cur != null; cur = cur.next.next) {
            if (cur.random != null) {
                cur.next.random = cur.random.next;
            }
        }
        Node ans = head.next;
        for (Node cur = head; cur != null;) {
            Node nxt = cur.next;
            if (nxt != null) {
                cur.next = nxt.next;
            }
            cur = nxt;
        }
        return ans;
    }
}

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) {
            return nullptr;
        }
        for (Node* cur = head; cur;) {
            Node* node = new Node(cur->val);
            node->next = cur->next;
            cur->next = node;
            cur = node->next;
        }
        for (Node* cur = head; cur; cur = cur->next->next) {
            if (cur->random) {
                cur->next->random = cur->random->next;
            }
        }
        Node* ans = head->next;
        for (Node* cur = head; cur;) {
            Node* nxt = cur->next;
            if (nxt) {
                cur->next = nxt->next;
            }
            cur = nxt;
        }
        return ans;
    }
};

Go

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

func copyRandomList(head *Node) *Node {
	if head == nil {
		return nil
	}
	for cur := head; cur != nil; {
		node := &Node{cur.Val, cur.Next, nil}
		cur.Next = node
		cur = node.Next
	}
	for cur := head; cur != nil; cur = cur.Next.Next {
		if cur.Random != nil {
			cur.Next.Random = cur.Random.Next
		}
	}
	ans := head.Next
	for cur := head; cur != nil; {
		nxt := cur.Next
		if nxt != nil {
			cur.Next = nxt.Next
		}
		cur = nxt
	}
	return ans
}

JavaScript

/**
 * // Definition for a Node.
 * function Node(val, next, random) {
 *    this.val = val;
 *    this.next = next;
 *    this.random = random;
 * };
 */

/**
 * @param {Node} head
 * @return {Node}
 */
var copyRandomList = function (head) {
    if (!head) {
        return null;
    }
    for (let cur = head; cur; ) {
        const node = new Node(cur.val, cur.next, null);
        cur.next = node;
        cur = node.next;
    }
    for (let cur = head; cur; cur = cur.next.next) {
        if (cur.random) {
            cur.next.random = cur.random.next;
        }
    }
    const ans = head.next;
    for (let cur = head; cur; ) {
        const nxt = cur.next;
        if (nxt) {
            cur.next = nxt.next;
        }
        cur = nxt;
    }
    return ans;
};

C#

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

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

public class Solution {
    public Node CopyRandomList(Node head) {
        if (head == null) {
            return null;
        }
        for (Node cur = head; cur != null; ) {
            Node node = new Node(cur.val, cur.next);
            cur.next = node;
            cur = node.next;
        }
        for (Node cur = head; cur != null; cur = cur.next.next) {
            if (cur.random != null) {
                cur.next.random = cur.random.next;
            }
        }
        Node ans = head.next;
        for (Node cur = head; cur != null; ) {
            Node nxt = cur.next;
            if (nxt != null) {
                cur.next = nxt.next;
            }
            cur = nxt;
        }
        return ans;
    }
}

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README.md

面试题 35. 复杂链表的复制

题目描述

解法

方法一：哈希表

Python3

Java

C++

Go

JavaScript

C#

Swift

方法二：拼接 + 拆分

Python3

Java

C++

Go

JavaScript

C#

Files

README.md

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README.md

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面试题 35. 复杂链表的复制

题目描述

解法

方法一：哈希表

Python3

Java

C++

Go

JavaScript

C#

Swift

方法二：拼接 + 拆分

Python3

Java

C++

Go

JavaScript

C#