133. Clone Graph

最新推荐文章于 2020-08-12 10:22:05 发布

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最新推荐文章于 2020-08-12 10:22:05 发布

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原文链接：http://www.cnblogs.com/duan-decode/p/9610731.html

本文详细介绍了一种用于克隆无向图的算法，通过深度优先搜索和哈希映射的方式，实现对图中每个节点及其邻居的完整复制。算法首先检查节点是否已复制，若未复制则创建新节点并递归复制其邻居，最终返回克隆后的图。

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 #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
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:

/**
 * 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==nullptr)
            return nullptr;
        //想到用map结构难一点
        unordered_map<const UndirectedGraphNode*, UndirectedGraphNode*> copied; //用map记录每一个node是否复制
        clone(node, copied);
        return copied[node];
    }
    static UndirectedGraphNode* clone(const UndirectedGraphNode* node, unordered_map<const UndirectedGraphNode*, UndirectedGraphNode*>& copied){
        if(copied.find(node)!=copied.end())
            return copied[node];    //有该节点直接返回（basecase：以自己为邻居）
        UndirectedGraphNode* new_node = new UndirectedGraphNode(node->label);   //map中没有该label值，申请新的节点存储原图节点
        copied[node] = new_node;
        for(auto nbr:node->neighbors)       //深度遍历该节点的邻居
            new_node->neighbors.push_back(clone(nbr,copied));
        return new_node;
    }
};