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() {
             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() {
             p1 = p1->next = p2->next;
             if(p1 == NULL) break;
             p2 = p2->next = p1->next;
         }
         return newhead;
     }
 };