本问题对应的 leetcode 原文链接：剑指 Offer 35. 复杂链表的复制
问题描述
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 。
代码实现
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;
}
}
时间复杂度：O(n)
额外空间复杂度：O(1)
Python
# Definition for a Node.
class Node:
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 not head:
return None
# 复制链表节点
cur = head
while cur:
next = cur.next
cur.next = Node(cur.val)
cur.next.next = next
cur = next
# 复制随机节点
cur = head
while cur:
curNew = cur.next
curNew.random = cur.random.next if cur.random else None
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:
cur.next = cur.next.next
cur = cur.next
curNew.next = cur.next if cur else None
curNew = curNew.next
return headNew
Python
Copy
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;
}
// 复制链表节点
Node* cur = head;
while (cur) {
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) {
Node* curNew = cur->next;
curNew->random = cur->random ? cur->random->next : nullptr;
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) {
cur->next = cur->next->next;
cur = cur->next;
curNew->next = cur ? cur->next : nullptr;
curNew = curNew->next;
}
return headNew;
}
};
C++
Copy
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 cur == nil {
curNew.Next = nil
} else {
curNew.Next = cur.Next
}
curNew = curNew.Next
}
return headNew
}
JS
/**
* // 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 == null) {
return null;
}
// 复制链表节点
let cur = head;
while(cur != null){
let 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){
let 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
let headNew = head.next;
cur = head;
let 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;
};
shuaidi
评论(2)
先拼接，再拆分