【剑指offer】【链表】35. 复杂链表的复制

题目链接：https://leetcode-cn.com/problems/fu-za-lian-biao-de-fu-zhi-lcof/

哈希

用哈希表做一个映射，hashmap的key是原先的节点，val是复制的节点。先初始化hashmap，然后分别next指针和random指针。
哈希表的作用：找到每个点的对应点（复制点）
时间复杂度O(n) 空间复杂度O(n)

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;
    
    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};
*/

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if(!head) return head;
        //使用哈希来做一个映射
        unordered_map<Node*, Node*> pos;

        pos[NULL] = NULL;
        //先复制正常的next节点，不复制random节点
        for(auto p = head; p; p = p -> next)
            pos[p] = new Node(p -> val);
        
        //复制指针的指向
        for(auto p = head; p; p = p -> next)
        {
            pos[p] -> next = pos[p -> next];
            pos[p] -> random = pos[p -> random];
        }
        return pos[head];
    }
};

迭代

先复制正常的next节点，不复制random节点。之后复制指针的指向
时间复杂度O(n) 空间复杂度O(1)

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;
    
    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};
*/
class Solution {
public:
    Node* copyRandomList(Node* head) {
        //在每个节点后面添加它的复制
        //复制next节点 复制的节点保存在原先节点的next指针处
        //新复制的节点的next指针的原先节点的next指针
        //原先： 1->2->3->4->5->null
        //now： 1->1->2->2->3->3->4->4->5->5->null
        for(auto p = head; p;)
        {
            auto np = new Node(p -> val);
            auto next = p -> next;
            p -> next = np;
            np -> next = next;
            p = next;
        }
	//复制random指针
        for(auto p = head; p; p = p -> next -> next)
            if(p -> random)
                p -> next -> random = p -> random -> next;
        //原链表不能修改 还原
        //遍历链表 分开两个链表
        auto dummy = new Node(-1);
        auto cur = dummy;
        for(auto p = head; p; p = p -> next)
        {
            cur -> next = p -> next;
            cur = cur -> next;
            p -> next = p -> next -> next;
        }
        return dummy -> next;
    }
};