133. Clone Graph

Problem

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:

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

Solution

/**
 * // Definition for a Node.
 * function Node(val, neighbors) {
 *    this.val = val === undefined ? 0 : val;
 *    this.neighbors = neighbors === undefined ? [] : neighbors;
 * };
 */

/**
 * @param {Node} node
 * @return {Node}
 */
var cloneGraph = function(node) {
    // Nodes we have already copied
    const visited = {};

    // DFS function to copy graph
    const dfs = (node) => {
        if (!node) return node;
        // If we have seen this node before, return it
        if (visited[node.val]!=null) return visited[node.val];

        // Create base for copied node
        const root = new Node(node.val);
        // Add this copied node to group of nodes we hav copied
        visited[node.val] = root;

        // Add copied neighbors to the current copied node
        node.neighbors.forEach(n => root.neighbors.push(dfs(n)))
        return root;
    }

    // Return new copied graph
    return dfs(node);
};

Explain:

不用管描述里的 # , 什么的，题意就是复制一个无向图，不要想复杂了。

注意有重复节点的话，存个引用就行，用哈希表判断是否重复。

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).