[LeetCode] Copy List with Random Pointer

最新推荐文章于 2019-10-21 22:50:54 发布

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本文介绍了一种解决复杂链表深拷贝问题的方法。该链表每个节点除了包含指向下一个节点的指针外，还有一个指向链表中任意节点的随机指针。通过三步走策略——克隆节点、复制随机指针、拆分链表，实现了对原链表的完整复制。

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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.
剑指OFFER上的题，中间 head->random 原来有可能是NULL,所以要添加一个判断，很坑。

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(!head)
            return NULL;
        cloneList(head);
        cloneRandomPtr(head);
        return removeClone(head);
    }
    void cloneList(RandomListNode* head)
    {
        while(head)
        {
            RandomListNode* toAdd=new RandomListNode(head->label);
            RandomListNode* pNext=head->next;
            toAdd->next=pNext;
            head->next=toAdd;
            head=pNext;
        }
    }
    void cloneRandomPtr(RandomListNode* head)
    {
        while(head)
        {
            if(head->random)
                head->next->random=head->random->next;
            head=head->next->next;
        }
    }
    RandomListNode* removeClone(RandomListNode* head)
    {
        RandomListNode guard(-1);
        RandomListNode* tail=&guard;
        while(head)
        {
            tail->next=head->next;
            tail=tail->next;
            head->next=head->next->next;
            head=head->next;
        }
        return guard.next;
    }
};