牛客OJ：复杂链表的复制

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/ShellDawn/article/details/89033262

哈希：

/*
struct RandomListNode {
    int label;
    struct RandomListNode *next, *random;
    RandomListNode(int x) :
            label(x), next(NULL), random(NULL) {
    }
};
*/
class Solution {
public:
   map<RandomListNode*,int> m;
vector<RandomListNode> v;

RandomListNode* solve(RandomListNode* p){
    if(p == NULL) return NULL;
    if(m[p] == 0){
        int L = v.size();
        RandomListNode t(p->label);
        v.push_back(t);
        m[p] = L+1; // m  :  L+1
        v[L].next = solve(p->next);
        v[L].random = solve(p->random);
        return &v[L];
    }else{
        return &v[m[p]-1];
    }
}

RandomListNode* Clone(RandomListNode* pHead)
{
    RandomListNode* ans = solve(pHead);
    return ans;
}
};