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

Author: dev-jonghoonpark

number-of-connected-components-in-an-undirected-graph/dev-jonghoonpark.md

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+- 문제
+  - 유료: https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/
+  - 무료: https://neetcode.io/problems/count-connected-components
+- 풀이: https://algorithm.jonghoonpark.com/2024/07/03/leetcode-323
+
+```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.