week10 mission number-of-connected-components-in-an-undirected-graph

_posts/2024-07-03-leetcode-323.md

@@ -0,0 +1,80 @@
+---
+layout: post
+title: (Leetcode) 323 - Number of Connected Components in an Undirected Graph
+categories: [스터디-알고리즘]
+tags: [자바, java, 리트코드, Leetcode, 알고리즘, tree, node, vertex, edge]
+date: 2024-07-02 12:30:00 +0900
+toc: true
+---
+
+기회가 되어 [달레님의 스터디](https://github.com/DaleStudy/leetcode-study)에 참여하여 시간이 될 때마다 한문제씩 풀어보고 있다.
+
+[https://neetcode.io/practice](https://neetcode.io/practice)
+
+---
+
+- 문제
+  - 유료: https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/
+  - 무료: https://neetcode.io/problems/count-connected-components
+
+## 내가 작성한 풀이
+
+```java
+public class Solution {
+    public int countComponents(int n, int[][] edges) {
+        Map<Integer, Vertex> vertexMap = new HashMap<>();
+        int lastGroupId = 0;
+
+        for (int[] edge : edges) {
+            Vertex v1 = vertexMap.getOrDefault(edge[0], new Vertex(edge[0]));
+            Vertex v2 = vertexMap.getOrDefault(edge[1], new Vertex(edge[1]));
+
+            v1.edges.add(v2);
+            v2.edges.add(v1);
+
+            vertexMap.put(edge[0], v1);
+            vertexMap.put(edge[1], v2);
+        }
+
+        for (int i = 0; i < n; i++) {
+            Vertex vertex = vertexMap.get(i);
+            if (vertex == null) {
+                lastGroupId++;
+            } else {
+                // 0 이 아닐 경우는 이미 탐색한 케이스므로 스킵
+                if (vertex.groupId == 0) {
+                    lastGroupId++;
+                    dfs(vertex, lastGroupId);
+                }
+            }
+        }
+
+        return lastGroupId;
+    }
+
+    public void dfs(Vertex vertex, int groupId) {
+        vertex.groupId = groupId;
+        for (Vertex connected : vertex.edges) {
+            if (connected.groupId == 0) {
+                dfs(connected, groupId);
+            }
+        }
+    }
+}
+
+class Vertex {
+    int id;
+    int groupId;
+    List<Vertex> edges;
+
+    public Vertex(int id) {
+        this.id = id;
+        this.edges = new ArrayList<>();
+    }
+}
+```
+
+### TC, SC
+
+Vertex 의 수를 V, Edge 의 수를 E 라고 하였을 때,
+시간 복잡도는 O(V + E), 공간 복잡도는 O(V + E) 이다.