Leetcode Copy List with Random Pointer

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

最先想到就是遍历列表生成新的列表，但没有对random域进行赋值。在遍历一次列表，然后进行random域赋值。但其存在一个问题，就是进行random赋值的时候需要耗费时间久，因为random域没有规律可循，需要进行遍历列表，也就是时间复杂度为O(N^2)。后面发现一个方法，就是通过对每个列表节点进行double，然后在进行random域的复制就变得有规律可循：

1.将列表内的节点全部复制，并连接在原节点后面

2.对新的节点的random域进行赋值，新节点.random = 原节点.random.next

3.将新节点从列表中抽离出新的列表

代码如下：

/** * Definition for singly-linked list with a random pointer. * struct RandomListNode { *     int label; *     RandomListNode *next, *random; *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {} * }; */class Solution {public:    RandomListNode *copyRandomList(RandomListNode *head) {        RandomListNode* next,*cur;        cur = head;        while(cur)        {            next = cur->next;            RandomListNode* tmp = new RandomListNode(cur->label);            tmp->next = next;            cur->next = tmp;            cur = next;        }                cur = head;        while(cur)        {            next = cur->next;            if(cur->random)                next->random = cur->random->next;            else                next->random = cur->random;            cur = next->next;        }                RandomListNode* result = new RandomListNode(0);        RandomListNode* copy,*copyCur=result;        cur = head;        while(cur)        {            next = cur->next->next;                        copyCur->next = cur->next;            copyCur = copyCur->next;                        cur->next = next;            cur = cur->next;        }        return result->next;            }};