leetcode每日一道（12）复杂链表的深拷贝

题目描述

现在有一个这样的链表：链表的每一个节点都附加了一个随机指针，随机指针可能指向链表中的任意一个节点或者指向空。
请对这个链表进行深拷贝。
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

三步走：

• 插入节点
• 复制随机指针
• 拆分链表

注意事项：这里的while可以用for来写，会更清爽一点。写了while就要注意每次的后移操作，一般报超时的错的时候，可能就是这个问题。总而言之，还是细节的考究！

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) return head;
        auto cur = head;
        RandomListNode* p;
        while(cur){
            p =new RandomListNode(cur->label);
            p->next = cur->next;
            cur->next = p;
            cur = p->next;
        }
        //随机指针的复制
        cur =head;
        while(cur){
            p = cur->next;
            p->random = (cur->random?cur->random->next:nullptr);
            cur = p->next;
        }
        cur =head;
        auto newhead = cur->next;
        while(cur){
            p=cur->next;
            cur = cur->next = p->next;
            p->next = (cur?cur->next:nullptr);
        }
        return newhead;
    }
};

for版本

RandomListNode *copyRandomList(RandomListNode *head) {
        RandomListNode *copy,*p;
        if (!head) return NULL;
        for(p=head;p;p=p->next){
            copy = new RandomListNode(p->label);
            copy->next = p->next; // insert new at old next
            p = p->next = copy;
        }
        for(p=head;p;p=copy->next){
            copy = p->next;          // copy random point
            copy->random = (p->random?p->random->next:NULL);
        }
        for(p=head,head=copy=p->next;p;){
            p = p->next = copy->next; // split list
            copy = copy->next = (p?p->next:NULL);
        }
        return head;
    }