Skip to main content

Graph - Example

Graph - Example


Leetcode 133. Clone Graph

Given a reference of a node in a connected undirected graph.

Return a deep copy (clone) of the graph.

Each node in the graph contains a value (int) and a list (List[Node]) of its neighbors.

class Node {

    public int val;

    public List<Node> neighbors;

}

Test case format:

For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with val == 1, the second node with val == 2, and so on. The graph is represented in the test case using an adjacency list.

An adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

Constraints:

The number of nodes in the graph is in the range [0, 100].

1 <= Node.val <= 100

Node.val is unique for each node.

There are no repeated edges and no self-loops in the graph.

The Graph is connected and all nodes can be visited starting from the given node.


Analysis:

This is a classic and also fundamental question in graph.

Before we try to clone a graph, let think about how to traversal a graph. Either dfs or bfs would work.

So we can add the cloning node during graph traversal.

The tricky part is how to maintain the edges between nodes, which connects the nodes together, or to form a graph.

One solution is to use a hash_map, which maps the node-to-node relationship between the original and the cloned graphs. With this map, it would be easy to re-construct the edges between the nodes in the cloned graph.

1. Only create the node when it does not exist, or is NULL;

2. Just need to add the neighbors nodes to each node in the cloned graph as that in the original graph.


See the code below:

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> neighbors;

    Node() {}

    Node(int _val, vector<Node*> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
};
*/
class Solution {
public:
    Node* cloneGraph(Node* node) {
        if(!node) return node;
        Node* copy = new Node(node->val, {});
        unordered_map<Node*, Node*> mp;
        mp[node] = copy;
        queue<Node*> q;
        q.push(node);
        while(q.size()) {
            auto t = q.front();
            q.pop();
            for(auto &a : t->neighbors) {
                if(!mp.count(a)) {
                    mp[a] = new Node(a->val, {});
                    q.push(a);
                }
                mp[t]->neighbors.push_back(mp[a]);
            }
        }
        return copy;
    }
};


Upper Layer

Comments

Popular posts from this blog

Sweep Line

Sweep (or scanning) line algorithm is very efficient for some specific questions involving discrete intervals. The intervals could be the lasting time of events, or the width of a building or an abstract square, etc. In the scanning line algorithm, we usually need to distinguish the start and the end of an interval. After the labeling of the starts and ends, we can sort them together based on the values of the starts and ends. Thus, if there are N intervals in total, we will have 2*N data points (since each interval will contribute 2). The sorting becomes the most time-consuming step, which is O(2N*log(2N) ~ O(N*logN). After the sorting, we usually can run a linear sweep for all the data points. If the data point is labeled as a starting point, it means a new interval is in the processing; when an ending time is reached, it means one of the interval has ended. In such direct way, we can easily figure out how many intervals are in the processes. Other related information can also be obt...

Bit Manipulation - Example

  Leetcode 136 Single Number Given a non-empty array of integers nums, every element appears twice except for one. Find that single one. You must implement a solution with a linear runtime complexity and use only constant extra space. Constraints: 1 <= nums.length <= 3 * 10^4 -3 * 10^4 <= nums[i] <= 3 * 10^4 Each element in the array appears twice except for one element which appears only once. Analysis: If there is no space limitation, this question can be solved by counting easily. But counting requires additional space. Here we can use xor (^) operation based on some interesting observations:  A^A = 0, here A is any number A^0 = A, here A is any number Since all the number appears twice except one, then all the number appear even numbers will be cancelled out, and only the number appears one time is left, which is what we want. See the code below: class Solution { public: int singleNumber(vector<int>& nums) { int res = 0; for(auto ...

Algorithm Advance Outline

Who wants to read these notes? 1. the one who wants to learn algorithm; 2. the one who has a planned interview in a short amount of time, such as one month later; 3. purely out of curiosity; 4. all the rest. The primary purpose for these posts is to help anyone who wants to learn algorithm in a short amount of time and be ready for coding interviews with tech companies, such as, Amazon, Facebook, Microsoft, etc. Before you start, you are expected to already know:  1. the fundamentals of at least one programming language; 2. the fundamentals of data structures. If you do not have these basics, it is better to "google and read" some intro docs, which should NOT take large amount of time. Of course, you can always learn whenever see a "unknown" data structure or line in the codes. Remember, "google and read" is always one of keys to learning. Common algorithms: 1. Recursion 2. Binary search 3. Dynamic programming 4. Breadth-first search 5. Depth-first search...