LeetCode 复制带随机指针的链表

题目链接：https://leetcode-cn.com/problems/copy-list-with-random-pointer/

题目大意

　　略。

分析

　　空间复杂度 O(1) 的做法非常开拓思维。

代码如下

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

    Node() {}

    Node(int _val, Node* _next, Node* _random) {
        val = _val;
        next = _next;
        random = _random;
    }
};
*/
class Solution {
public:
    Node* copyRandomList(Node* head) {
        if(head == NULL) return NULL;
        
        Node *p1 = head, *p2, *newhead;
        
        // 在原链表中交替嵌入新节点
        while(p1 != NULL) {
            Node *t = new Node();
            
            t->next = p1->next;
            t->val = p1->val;
            p1->next = t;
            p1 = t->next;
        }
        
        newhead = head->next;
        p1 = head;
        p2 = newhead;
        // 拷贝random
        while(1) {
            if(p1->random == NULL) p2->random = NULL;
            else p2->random = p1->random->next;
            p1 = p2->next;
            if(p1 == NULL) break;
            p2 = p1->next;
        }
        
        p1 = head;
        p2 = newhead;
        // 分离
        while(1) {
            p1 = p1->next = p2->next;
            if(p1 == NULL) break;
            p2 = p2->next = p1->next;
        }
        return newhead;
    }
};