#133 Clone Graph

同步發佈於Github repo

題目難度：Medium

題目敘述：

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
         / \
         \_/

題目解答：

/**
 * Definition for undirected graph.
 * function UndirectedGraphNode(label) {
 *     this.label = label;
 *     this.neighbors = [];   // Array of UndirectedGraphNode
 * }
 */

/**
 * @param {UndirectedGraphNode} graph
 * @return {UndirectedGraphNode}
 */
const dfs = (node, visited) => {
  let newNode = visited[node.label] ? visited[node.label] : new UndirectedGraphNode(node.label);
  visited[node.label] = newNode;
        
  for(let i = 0; i < node.neighbors.length; i++) {
    const newNeighbor = visited[node.neighbors[i].label] ? visited[node.neighbors[i].label] : dfs(node.neighbors[i], visited);
    newNode.neighbors.push(newNeighbor);
  }

  return newNode; 
}; 
 
const cloneGraph = graph => {
  let visited = {};
    
  if(graph === null){
    return graph;
  } else {
    return dfs(graph, visited);
  }
};

題目連結：133. Clone Graph

解題說明：

今天是第 18 天啦～ 終於要度過 graph 篇了！

今天的題目題意很簡單，給你一個無向圖，要我們歷遍整個圖，並複製一個出來，
這題是考對 graph 資料結構的歷遍方法而已，
我們使用 DFS 來歷遍圖，並且用 JavaScript 的 Object 實作 HashMap 的用途，負責記錄走過的點，
這樣就可以了～

解題步驟：

• 步驟 1.
  我們用 visited 這個 Object 來存我們走過的點，但最後不是回傳這個變數，這只是存在過程中走過的點而已。

let visited = {};
    
if(graph === null){
  return graph;
} else {
  return dfs(graph, visited);
}

• 步驟 2.
  這步驟是最主要的 DFS 歷遍的 function，
  newNode 是用來存每一次迭代時最新的節點，第一次進來則是 new UndirectedGraphNode(node.label)，
  然後把它放進 visited 裡面記錄起來，

  最後只要在下面的 for loop 裡面每次用 DFS 去檢查該節點有沒有相鄰的節點，如果有的話就把它放進 neighbors 裡面，
  完成！

const dfs = (node, visited) => {
  let newNode = visited[node.label] ? visited[node.label] : new UndirectedGraphNode(node.label);
  visited[node.label] = newNode;
        
  for(let i = 0; i < node.neighbors.length; i++) {
    const newNeighbor = visited[node.neighbors[i].label] ? visited[node.neighbors[i].label] : dfs(node.neighbors[i], visited);
    newNode.neighbors.push(newNeighbor);
  }

  return newNode; 
};