Clone Graph -- LeetCode

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。