复制带随机指针的链表(LeetCode中级篇138题)_leetcode 复制带随机指针的链表-优快云博客

本文链接：https://blog.youkuaiyun.com/ADream__/article/details/85008367

本文详细介绍了一种链表深度拷贝的实现方法，针对含有随机指针的单链表，通过三个步骤：复制所有节点、连接random指针、提取复制链表，实现了原链表的完全克隆。

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给定一个链表，每个节点包含一个额外增加的随机指针，该指针可以指向链表中的任何节点或空节点。

要求返回这个链表的深度拷贝。

/**
 * Definition for singly-linked list with a random pointer.
 * struct RandomListNode {
 *     int label;
 *     struct RandomListNode *next;
 *     struct RandomListNode *random;
 * };
 */
typedef struct RandomListNode RLNote;
struct RandomListNode *copyRandomList(struct RandomListNode *head) {
    //第一步，先复制链表中的所有结点
    RLNote *cur = head;
    RLNote *next;
    while(cur)
    {
        next = cur->next;
        RLNote *copy =(RLNote *)malloc(sizeof(RLNote));
        copy->label = cur->label;
        cur->next = copy;
        copy->next = next;
        cur = next;
    }
    //第二步，将random连接起来
    cur = head;
    while(cur)
    {
        RLNote* copy = cur->next;
        if(cur->random)
            copy->random = cur->random->next;
        else
            copy->random = NULL;
        cur = copy->next;
    }
    //第三步，将copy提取出来
    RLNote* copyhead,* tail ;
    copyhead = tail = (RLNote*)malloc(sizeof(RLNote));
    cur = head;
    while(cur)
    {
        RLNote* copy = cur->next;
        RLNote* next = copy->next;
        
        tail -> next = copy;
        tail = copy;
        //链接原链表的所有next
        cur -> next = next;  
        //循环下一个时将cur指向next
        cur = next; 
    }
    RLNote* newnode = copyhead->next;
    free(copyhead);
    return newnode;
}