leetcode 138. 随机链表的复制

Problem: 138. 随机链表的复制

方案

使用列表来保存，并按照index来new节点，根据random指向地址对应到index

复杂度

时间复杂度:

添加时间复杂度, 示例： $O(n)$

空间复杂度:

添加空间复杂度, 示例： $O(n)$

Code

/*
// 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==nullptr) return nullptr;
        Node *p0 = head, *p1, *p2, *p3;
        vector<pair<int, Node *>> tr;
        vector<Node *> trc, trr;
        vector<int> ump;
        while(p0!=nullptr) {
            tr.push_back({p0->val, p0->random});
            trc.push_back(p0);
            p0 = p0->next;
        }
        tr.push_back({-1, nullptr});
        trc.push_back(nullptr);
        for(int i = 0 ; i < tr.size(); i++) {
            p1 = tr[i].second;
            for(int j = 0 ; j < trc.size(); j++) {
                if(p1==trc[j]) {
                    ump.push_back(j);
                }
            }
        }
        p1 = p2 = new Node(tr[0].first);
        trr.push_back(p1);
        for(int i = 1 ; i < tr.size(); i++) {
            if(i!=tr.size()-1) p3 = new Node(tr[i].first);
            else p3 = nullptr;
            trr.push_back(p3);
            p2->next = p3;
            p2 = p2->next;
        }
        for(int i = 0 ; i < trr.size()-1; i++) {
            trr[i]->random = trr[ump[i]];
        }
        return p1;
    }
};
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