《中英双解》leetCode Find if path exists in graph

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 start to vertex end.

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

有一个具有 n个顶点的 双向 图，其中每个顶点标记从 0 到 n - 1（包含 0 和 n - 1）。图中的边用一个二维整数数组 edges 表示，其中 edges[i] = [ui, vi] 表示顶点 ui 和顶点 vi 之间的双向边。 每个顶点对由 最多一条 边连接，并且没有顶点存在与自身相连的边。

请你确定是否存在从顶点 start 开始，到顶点 end 结束的 有效路径 。

给你数组 edges 和整数 n、start和end，如果从 start 到 end 存在 有效路径 ，则返回 true，否则返回 false 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-if-path-exists-in-graph
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Example 1:

Input: n = 3, edges = [[0,1],[1,2],[2,0]], start = 0, end = 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]], start = 0, end = 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 <= start, end <= n - 1
• There are no duplicate edges.
• There are no self edges.

BFS

class Solution {
    public boolean validPath(int n, int[][] edges, int start, int end) {
        HashMap<Integer, List<Integer>> graph = new HashMap<Integer,List<Integer>>();
        for (int i= 0;i <edges.length;i++)
        {
            if(graph.containsKey(edges[i][0]))
            {
                graph.get(edges[i][0]).add(edges[i][1]);
            }
            else
            {
                List<Integer> temp = new ArrayList<Integer>();
                temp.add(edges[i][1]);
                graph.put(edges[i][0],temp);
            }
            if(graph.containsKey(edges[i][1]))
            {
                graph.get(edges[i][1]).add(edges[i][0]);
            }
            else
            {
                List<Integer> temp2 = new ArrayList<Integer>();
                temp2.add(edges[i][0]);
                graph.put(edges[i][1],temp2);
            }
        }
        int[] visited = new int[n];
        Queue<Integer> queue = new LinkedList<Integer>();
        queue.add(start);
        visited[start] = 1;
        while(!queue.isEmpty())
        {
            int top = queue.remove();
            if(graph.containsKey(top))
            {
                for(int k=0;k < graph.get(top).size();k++)
                {
                    if(visited[graph.get(top).get(k)] != 1)
                    {
                        queue.add(graph.get(top).get(k));
                        visited[graph.get(top).get(k)] = 1;
                    }
                }
            }
        }
        return visited[end] == 1;
        
    }
}

 DFS

class Solution {
    public boolean validPath(int n, int[][] edges, int start, int end) {
       boolean[] visited = new boolean[n + 1];
       boolean newVisited = true;
       visited[start] = true;

       while(!visited[end] && newVisited){
           newVisited = false;
           for(int[] edge : edges){
              if(visited[edge[0]] && !visited[edge[1]]){
                  visited[edge[1]] = true;
                  newVisited = true;
              }
              if(!visited[edge[0]] && visited[edge[1]]){
                  visited[edge[0]] = true;
                  newVisited = true;
              }
           }
       }
       return visited[end];
   }
}

标签：end,双解,exists,int,graph,start,edges,visited
来源： https://blog.csdn.net/weixin_46272350/article/details/121595221