[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.

思路：首先将原链表复制一个新节点并连接到所复制的节点后面，然后在这样的链表上链接复制节点的random指针。最后分拆链表

　　　时间复杂度O(n),空间复杂度O(1)

/**
 * 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) {
        if (head == NULL) return head;
        
        //复制节点
        for (RandomListNode *curr = head; curr != NULL;) {
            RandomListNode *pter = new RandomListNode(curr->label);
            pter->next = curr->next;
            curr->next = pter;
            curr = pter->next;
        }
        
        //链接random指针
        for(RandomListNode *curr = head; curr != NULL;) {
            if (curr->random != NULL) {
                curr->next->random = curr->random->next;
            }
            curr = curr->next->next;
        }
        
        //分拆链表
        RandomListNode *new_head = head->next;
        for (RandomListNode *curr = head, **new_curr = &new_head; curr != NULL;) {
            *new_curr = curr->next;
            curr->next = curr->next->next;
            
            curr = curr->next;
            
            new_curr = &((*new_curr)->next);
        }
        
        return new_head;
    }
};