-
Notifications
You must be signed in to change notification settings - Fork 90
/
Clone_Graph.js
100 lines (73 loc) · 2.77 KB
/
Clone_Graph.js
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
/*
Clone Graph
https://leetcode.com/problems/clone-graph/description/
Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label (int) and a list (List[UndirectedGraphNode]) of its neighbors. There is an edge between the given node and each of the nodes in its neighbors.
OJ's undirected graph serialization (so you can understand error output):
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
/ \
\_/
Note: The information about the tree serialization is only meant so that you can understand error output if you get a wrong answer.
You don't need to understand the serialization to solve the problem.
*/
/**
* Definition for undirected graph.
* function UndirectedGraphNode(label) {
* this.label = label;
* this.neighbors = []; // Array of UndirectedGraphNode
* }
*/
class UndirectedGraphNode {} //TODO: Define me
// SOLUTION 1 Using DFS
/**
* @param {UndirectedGraphNode} graph
* @return {UndirectedGraphNode}
*/
var cloneGraph = function (graph) {
if (!graph) return graph;
return dfs(graph, {});
};
var dfs = function (graph, visited) {
if (visited[graph.label]) return visited[graph.label];
var newNode = new UndirectedGraphNode(graph.label);
visited[newNode.label] = newNode;
for (var i = 0; i < graph.neighbors.length; i++) {
const neighbor = dfs(graph.neighbors[i], visited);
newNode.neighbors.push(neighbor);
}
return newNode;
};
// SOLUTION 2 Using DFS
var cloneGraphBFS = function (graph) {
if (graph === null) return graph;
var visitedMap = {};
var queue = [graph];
var copyReturn = new UndirectedGraphNode(graph.label);
visitedMap[graph.label] = copyReturn;
while (queue.length > 0) {
var node = queue.shift();
var nodeCopied = visitedMap[node.label];
for (var i = 0; i < node.neighbors.length; i++) {
var neighbor = node.neighbors[i];
if (!visitedMap[neighbor.label]) {
var copyNeighbor = new UndirectedGraphNode(neighbor.label);
visitedMap[neighbor.label] = copyNeighbor;
queue.push(neighbor);
}
nodeCopied.neighbors.push(visitedMap[neighbor.label]);
}
}
return copyReturn;
};
module.exports.cloneGraph = cloneGraph;
module.exports.cloneGraphBFS = cloneGraphBFS;