力扣算法——133.CloneGraph【M】

Given a reference of a node in a connected undirected graph, return a deep copy (clone) of the graph. Each node in the graph contains a val (int) and a list (List[Node]) of its neighbors.

Example:

Input:
{"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1}

Explanation:
Node 1's value is 1, and it has two neighbors: Node 2 and 4.
Node 2's value is 2, and it has two neighbors: Node 1 and 3.
Node 3's value is 3, and it has two neighbors: Node 2 and 4.
Node 4's value is 4, and it has two neighbors: Node 1 and 3.

Note:

1. The number of nodes will be between 1 and 100.
2. The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
3. Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
4. You must return the copy of the given node as a reference to the cloned graph.

Soulution:

　　使用DFS和BFS即可

　　两种方法在leetcode中都能通过，但在牛客中，BFS使用的而外空间超出限制，所以只能使用DFS

　　

struct UndirectedGraphNode {
    int label;
    vector<UndirectedGraphNode *> neighbors;
    UndirectedGraphNode(int x) : label(x) {};
};

//DFS
class Solution01 {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        unordered_map<UndirectedGraphNode*, UndirectedGraphNode*>mapR;
        return helper(node, mapR);
    }
    UndirectedGraphNode* helper(UndirectedGraphNode*node, unordered_map<UndirectedGraphNode*, UndirectedGraphNode*>&mapR)
    {
        if (node == nullptr)return nullptr;
        if (mapR.find(node) != mapR.end())//旧节点
            return mapR[node];
        UndirectedGraphNode* res = new UndirectedGraphNode(node->label);//新节点
        mapR[node] = res;
        for (auto a : node->neighbors)
            res->neighbors.push_back(helper(a, mapR));
        return res;
    }
};
//BFS
class Solution02 {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        queue<UndirectedGraphNode*>qNode;
        qNode.push(node);
        unordered_map<UndirectedGraphNode*, UndirectedGraphNode*>mapR;
        UndirectedGraphNode *res = new UndirectedGraphNode(node->label);
        mapR[node] = res;
        while (!qNode.empty())
        {
            UndirectedGraphNode *p = qNode.front();
            qNode.pop();
            for (auto a : p->neighbors)
            {
                if (mapR.find(a) == mapR.end())//新节点
                {
                    qNode.push(a);//加入
                    mapR[a] = new UndirectedGraphNode(a->label);//新节点
                }
                mapR[p]->neighbors.push_back(mapR[a]);//连接节点
            }
        }
        return res;        
    }
};