133 Clone Graph

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
         / \
         \_/

原题链接: http://oj.leetcode.com/problems/clone-graph/ 

这道题是LeetCode中为数不多的关于图的题目，不过这道题还是比较基础，就是考察图非常经典的方法：深度优先搜索和广度优先搜索。这道题用两种方法都可以解决，因为只是一个图的复制，用哪种遍历方式都可以。具体细节就不多说了，因为两种方法太常见了。这里恰好可以用旧结点和新结点的HashMap来做visited的记录。下面是广度优先搜索的代码： 

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node==null)
        return null;
    LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    UndirectedGraphNode copy = new UndirectedGraphNode(node.label);
    map.put(node,copy);
    queue.offer(node);
    while(!queue.isEmpty())
    {
        UndirectedGraphNode cur = queue.poll();
        for(int i=0;i<cur.neighbors.size();i++)
        {
            if(!map.containsKey(cur.neighbors.get(i)))
            {
                copy = new UndirectedGraphNode(cur.neighbors.get(i).label);
                map.put(cur.neighbors.get(i),copy);
                queue.offer(cur.neighbors.get(i));
            }
            map.get(cur).neighbors.add(map.get(cur.neighbors.get(i)));
        }
    }
    return map.get(node);
}

深度优先搜索的代码如下：

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node == null)
        return null;
    LinkedList<UndirectedGraphNode> stack = new LinkedList<UndirectedGraphNode>();
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    stack.push(node);
    UndirectedGraphNode copy = new UndirectedGraphNode(node.label);
    map.put(node,copy);
    while(!stack.isEmpty())
    {
        UndirectedGraphNode cur = stack.pop();
        for(int i=0;i<cur.neighbors.size();i++)
        {
            if(!map.containsKey(cur.neighbors.get(i)))
            {
                copy = new UndirectedGraphNode(cur.neighbors.get(i).label);
                map.put(cur.neighbors.get(i),copy);
                stack.push(cur.neighbors.get(i));
            }
            map.get(cur).neighbors.add(map.get(cur.neighbors.get(i)));
        }
    }
    return map.get(node);
}

当然深度优先搜索也可以用递归来实现，代码如下：

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node == null)
        return null;
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    UndirectedGraphNode copy = new UndirectedGraphNode(node.label);
    map.put(node,copy);
    helper(node,map);
    return copy;
}
private void helper(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map)
{
    for(int i=0;i<node.neighbors.size();i++)
    {
        UndirectedGraphNode cur = node.neighbors.get(i);
        if(!map.containsKey(cur))
        {
            UndirectedGraphNode copy = new UndirectedGraphNode(cur.label);
            map.put(cur,copy);
            helper(cur,map);
        }
        map.get(node).neighbors.add(map.get(cur));
    }
}

这
几种方法的时间复杂度都是O(n)（每个结点访问一次），而空间复杂度则是栈或者队列的大小加上HashMap的大小，也不会超过O(n)。图的两种遍历
方式是比较经典的问题了，虽然在面试中出现的不多，但是还是有可能出现的，而且如果出现了就必须做好，所以大家还是得好好掌握哈。

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if (node == NULL) return NULL;
        
        unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> mapper;
        UndirectedGraphNode *newNode = new UndirectedGraphNode(node->label);
        mapper[node] = newNode;
        queue<UndirectedGraphNode*> q;
        q.push(node);
        
        while (!q.empty()) {
            UndirectedGraphNode *cur = q.front();
            q.pop();
            for (int i = 0; i < cur->neighbors.size(); i++) {
                UndirectedGraphNode *temp = cur->neighbors[i];
                if (mapper.find(temp) == mapper.end()) {
                    mapper[temp] = new UndirectedGraphNode(temp->label);
                    q.push(temp);
                }
                mapper[cur]->neighbors.push_back(mapper[temp]);
            }
        }

        return newNode;        
    }
};