[LeetCode][JavaScript]Clone Graph

Clone Graph 

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 #.

  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
         / \
         \_/

https://leetcode.com/problems/clone-graph/







节点是唯一的，就是要注意像例子中这样从2到2的环，有个test case是{0,0,0}

DFS递归解法：
 1 /**
 2  * Definition for undirected graph.
 3  * function UndirectedGraphNode(label) {
 4  *     this.label = label;
 5  *     this.neighbors = [];   // Array of UndirectedGraphNode
 6  * }
 7  */
 8 /**
 9  * @param {UndirectedGraphNode} graph
10  * @return {UndirectedGraphNode}
11  */
12 var cloneGraph = function(graph) {
13     var visited = {};
14     if(graph === null){
15         return null;
16     }else{
17         return dfs(graph);
18     }
19 
20     function dfs(node){
21         var newNode = null;
22         if(visited[node.label]){
23             newNode = visited[node.label];
24         }else{
25             newNode = new UndirectedGraphNode(node.label);
26             visited[node.label] = newNode;
27         }
28         for(var i = 0; i < node.neighbors.length; i++){
29             if(node.neighbors[i].label !== node.label){
30                 if(!visited[node.neighbors[i].label]){
31                     newNode.neighbors.push(dfs(node.neighbors[i]));
32                 }else{
33                     newNode.neighbors.push(visited[node.neighbors[i].label]);
34                 }
35             }else{
36                 newNode.neighbors.push(visited[node.label]);
37             }
38         }
39         return newNode; 
40     }   
41 };

 


BFS非递归解法:

 1 /**
 2  * Definition for undirected graph.
 3  * function UndirectedGraphNode(label) {
 4  *     this.label = label;
 5  *     this.neighbors = [];   // Array of UndirectedGraphNode
 6  * }
 7  */
 8 
 9 /**
10  * @param {UndirectedGraphNode} graph
11  * @return {UndirectedGraphNode}
12  */
13 var cloneGraph = function(graph) {
14     var visited = {};
15     var queue = [];
16     if(graph === null){
17         return null;
18     }else{
19         visited[graph.label] = new UndirectedGraphNode(graph.label);
20         queue.push(graph);
21         bfs();
22         return visited[graph.label];
23     }
24 
25     function bfs(){
26         while(queue.length > 0){
27             var node = queue.shift();
28             var newNode = visited[node.label];
29             for(var i = 0; i < node.neighbors.length; i++){
30                 if(node.neighbors[i].label !== node.label){
31                     var neighbor = node.neighbors[i];
32                     var newNeighbor = null;
33                     if(!visited[neighbor.label]){
34                         newNeighbor = new UndirectedGraphNode(neighbor.label);
35                         visited[neighbor.label] = newNeighbor;
36                         queue.push(neighbor);
37                     }else{
38                         newNeighbor = visited[neighbor.label];
39                     }
40                     newNode.neighbors.push(newNeighbor);
41                 }else{
42                     newNode.neighbors.push(visited[node.label]);
43                 }
44             }
45         }
46     } 
47 };

 

posted @ 2015-05-20 13:20  `Liok  阅读(761)  评论(2)  编辑  收藏  举报