[leetcode]133. Clone Graph

题目链接：https://leetcode.com/problems/clone-graph/tabs/description

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ's undirected graph serialization:

Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1
      / \
     /   \
    0 --- 2
         / \
         \_/

In order to clone a graph, you need to have a copy of each node in the original graph. Well, you may not have too many ideas about it. Let's do an example.

Suppose we are given a graph {0, 1 # 1, 0}. We know that the graph has two nodes 0 and 1 and they are connected to each other.

We now start from 0. We make a copy of 0. Then we check 0's neighbors and we see 1. We make a copy of 1 and we add the copy to the neighbors of the copy of 0. Now the cloned graph is {0 (copy), 1 (copy)}. Then we visit 1. We make a copy of 1... well, wait, why do we make another copy of it? We already have one! Note that if you make a new copy of the node, these copies are not the same and the graph structure will be wrong! This is just what I mean by "the most tricky part of this problem". In fact, we need to maintain a mapping from each node to its copy. If the node has an existed copy, we simply use it. So in the above example, the remaining process is that we visit the copy of 1 and add the copy of 0 to its neighbors and the cloned graph is eventually {0 (copy), 1 (copy) # 1 (copy), 0 (copy)}.

方法一：（BFS）

class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if (!node) return NULL;
        UndirectedGraphNode* copy = new UndirectedGraphNode(node -> label);
        mp[node] = copy;
        queue<UndirectedGraphNode*> toVisit;
        toVisit.push(node);
        while (!toVisit.empty()) {
            UndirectedGraphNode* cur = toVisit.front();
            toVisit.pop();
            for (UndirectedGraphNode* neigh : cur -> neighbors) {
                if (mp.find(neigh) == mp.end()) {
                    UndirectedGraphNode* neigh_copy = new UndirectedGraphNode(neigh -> label);
                    mp[neigh] = neigh_copy;
                    toVisit.push(neigh);
                }
                mp[cur] -> neighbors.push_back(mp[neigh]);
            }
        }
        return copy; 
    }
private:
    unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> mp;
};

方法二：（DFS）

class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(!node) return NULL;
        if(mp.find(node)==mp.end())
        {
            mp[node]=new UndirectedGraphNode(node->label);
            for(UndirectedGraphNode* neigh:node->neighbors)
                mp[node]->neighbors.push_back(cloneGraph(neigh));
        }
        return mp[node];
    }
private:
    unordered_map<UndirectedGraphNode*,UndirectedGraphNode*> mp;
};