【Leetcode】【Hard】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.

解题思路：

本题要求新建一个链表，使其完全等于输入链表，相当于对输入链表深复制。

题目的难点在于，原链表中有random指针，指向原链表对应的结点。当新链表建立时，需要对新结点的random指针赋值，使其指向新链表中的对应结点。这样就不能简单的复制地址，需要在新结点建立后，再找到对应正确的地址。暴力解法的时间复杂度为o(n2)。

o(n)的解法：

将每一个新结点，建立在旧结点的后面，形成：old1->new1->old2->new2->...oldN->newN->...；o(n)

建立后，重新遍历这个加长链表，new1->random = old1->random->next；o(n)

最后将新旧链表拆分；o(n)

最终时间复杂度为o(n)，空间复杂度为o(1)

代码：

/**
 * 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;
        
        RandomListNode* oldNode = head;
        RandomListNode* nodeLeft = NULL;
        RandomListNode* newNode = NULL;
        RandomListNode* newList = NULL;
        
        while (oldNode) {
            nodeLeft = oldNode->next;
            oldNode->next = new RandomListNode(oldNode->label);
            oldNode->next->next = nodeLeft;
            oldNode = nodeLeft;
        }
        
        oldNode = head;
        newList = head->next;
        
        while (oldNode) {
            newNode = oldNode->next;
            if (oldNode->random == NULL)
                newNode->random = NULL;
            else 
                newNode->random = oldNode->random->next;
            oldNode = newNode->next;
        }
        
        oldNode = head;
        while (oldNode) {
            newNode = oldNode->next;
            oldNode->next = newNode->next;
            oldNode = oldNode->next;
            if (oldNode)
                newNode->next = oldNode->next;
        }
        
        return newList;
    }
};