码迷,mamicode.com
首页 > 其他好文 > 详细

leetcode Clone Graph

时间:2014-12-10 22:31:27      阅读:228      评论:0      收藏:0      [点我收藏+]

标签:des   style   blog   http   io   ar   color   os   sp   

复制一个无向图。图的结构时有一个label,一个vector存和他想接的节点。可以自循环,就是vector中可以存在自己。例如:

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
         /          \_/
思路:一开始,我就想到用BFS,但是需要判断是否已经处理过该节点,因为存在循环。避免死循环就要用一个可以O1时间访问否值是否存在的东东来存已经访问过的节点,那就是map了,或者是set,我用set存访问过的,然后两个queue来同步的走,及把node复制到另一个root中,然后que和que2分别表示需要继续BFS处理的后序节点。非迭代的:
bubuko.com,布布扣
/**
 * 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) return node;
        //if ((node -> neighbors).size() == 0) return node;
        queue<UndirectedGraphNode *> que, que2;
        que.push(node);
        UndirectedGraphNode *root = new UndirectedGraphNode(node -> label), *tmp, *subnode, *tmp2;
        unordered_set<UndirectedGraphNode*> uset;
        que2.push(root);
        while(!que.empty())
        {
            tmp = que.front();
            tmp2 = que2.front();
            uset.insert(tmp);
            que.pop();
            que2.pop();
            for (int i = 0; i < tmp -> neighbors.size(); ++i)
            {
                if (tmp -> neighbors[i] == tmp)
                    subnode = tmp2;
                else
                    subnode = new UndirectedGraphNode(tmp -> neighbors[i] -> label);
                tmp2 -> neighbors.push_back(subnode);
                if (uset.count(tmp -> neighbors[i]) == 0)
                {
                    que.push(tmp -> neighbors[i]);
                    que2.push(subnode);
                }
            }
        }
        return root;
    }
};
View Code

但是超时了,可能最近感冒脑子不灵活了哎。

不知道为什么上面用两个队列会超时啊。换成map的映射的话就不会超时:

unordered_map(UndirectedGraphNode *, UndirectedGraphNode *) copied; 指key对应的node是否在value对应的node中复制。

因为是BFS要用一个que来存自己已经复制,但是他的neighbors还没有复制的节点。那么初始化que就要push一个node,而copied[node]就是赋值为node->label对应的新节点。

然后根据que是否为空处理。

/**
 * 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) return node;
        unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> copied;
        queue<UndirectedGraphNode *> que;
        que.push(node);
        copied[node] = new UndirectedGraphNode(node -> label);
        
        while(!que.empty())
        {
            UndirectedGraphNode *cur = que.front();
            que.pop();
            for (int i = 0; i < cur -> neighbors.size(); ++i)
            {
                if (copied.count(cur->neighbors[i]))
                    copied[cur] -> neighbors.push_back(copied[cur -> neighbors[i]]);
                else
                {
                    UndirectedGraphNode *new_node = new UndirectedGraphNode(cur -> neighbors[i] -> label);
                    copied[cur -> neighbors[i]] = new_node;
                    copied[cur] -> neighbors.push_back(new_node);
                    que.push(cur -> neighbors[i]);
                }
            }
        }
        return copied[node];
    }
};

 这有DFS

leetcode Clone Graph

标签:des   style   blog   http   io   ar   color   os   sp   

原文地址:http://www.cnblogs.com/higerzhang/p/4156373.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!