99클럽 코테 스터디 25일차 TIL Find if Path Exists in Graph

# 오늘의 학습 키워드 

그래프

 

# 오늘의 문제 

https://leetcode.com/problems/find-if-path-exists-in-graph/description/

There is a bi-directional graph with n vertices, where each vertex is labeled from 0 to n - 1 (inclusive). The edges in the graph are represented as a 2D integer array edges, where each edges[i] = [ui, vi] denotes a bi-directional edge between vertex ui and vertex vi. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself.

You want to determine if there is a valid path that exists from vertex source to vertex destination.

Given edges and the integers n, source, and destination, return true if there is a valid path from source to destination, or false otherwise.

 

Example 1:

 

Input: n = 3, edges = [[0,1],[1,2],[2,0]], source = 0, destination = 2

Output: true

Explanation: There are two paths from vertex 0 to vertex 2:

- 0 → 1 → 2

- 0 → 2

Example 2:

 

Input: n = 6, edges = [[0,1],[0,2],[3,5],[5,4],[4,3]], source = 0, destination = 5

Output: false

Explanation: There is no path from vertex 0 to vertex 5.

 

Constraints:

• 1 <= n <= 2 * 105
• 0 <= edges.length <= 2 * 105
• edges[i].length == 2
• 0 <= ui, vi <= n - 1
• ui != vi
• 0 <= source, destination <= n - 1
• There are no duplicate edges.
• There are no self edges.

 

# 나의 풀이방식 

class Solution {
    public boolean validPath(int n, int[][] edges, int source, int destination) {
        Map<Integer,List<Integer>> graph = new HashMap<>();
        boolean[] visited = new boolean[n]; 

        for(int[] edge : edges){
            int a = edge[0];
            int b = edge[1];
            graph.computeIfAbsent(a,k->new ArrayList<Integer>()).add(b);
            graph.computeIfAbsent(b,k->new ArrayList<Integer>()).add(a);
        }

        return dfs(graph, visited, source, destination);


    }


    private boolean dfs(Map<Integer, List<Integer>> graph, boolean[] visited, int start, int end){
        if(start == end) return true;

        if(!visited[start]){
            visited[start] = true;
            for(int next : graph.get(start)){
                if(dfs(graph,visited,next,end)){
                    return true;
                }
            }
        }
        return false;
    }
}

# 오늘의 공부 

시간 복잡도 = O(V+E) 

(V=정점의 개수, E=간선의 개수)

 

# 오늘의 회고 

 

 그래프 문제 좀더 풀어봐야겠다....