> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/graph-search/815.bus-routes.md).

# 815.Bus-Routes

## 815. Bus Routes

## 题目地址

<https://leetcode.com/problems/bus-routes/>

## 题目描述

```
We have a list of bus routes. Each routes[i] is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever.

We start at bus stop S (initially not on a bus), and we want to go to bus stop T. Travelling by buses only, what is the least number of buses we must take to reach our destination? Return -1 if it is not possible.

Example:
Input: 
routes = [[1, 2, 7], [3, 6, 7]]
S = 1
T = 6
Output: 2
Explanation: 
The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.

Note:
1 <= routes.length <= 500.
1 <= routes[i].length <= 500.
0 <= routes[i][j] < 10 ^ 6.
```

## 代码

### Approach 1: BFS

```java
class Solution {
  public int numBusesToDestination(int[][] routes, int S, int T) {
    HashSet<Integer> visited = new HashSet<>();
    Queue<Integer> q = new LinkedList<>();
    HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();
    int ret = 0; 

    if (S == T) return 0; 

    for (int i = 0; i < routes.length; i++) {
      for (int j = 0; j < routes[i].length; j++) {
        ArrayList<Integer> buses = map.getOrDefault(routes[i][j], new ArrayList<>());
        buses.add(i);
        map.put(routes[i][j], buses);                
      }       
    }

    q.offer(S); 
    while (!q.isEmpty()) {
      int len = q.size();
      ret++;
      for (int i = 0; i < len; i++) {
        int cur = q.poll();
        ArrayList<Integer> buses = map.get(cur);
        for (int bus: buses) {
          if (visited.contains(bus)) continue;
          visited.add(bus);
          for (int j = 0; j < routes[bus].length; j++) {
            if (routes[bus][j] == T) return ret;
            q.offer(routes[bus][j]);  
          }
        }
      }
    }
    return -1;
  }
}
```

### Approach #2 BFS

```java
public int numBusesToDestination(int[][] routes, int S, int T) {
    int n = routes.length;
    HashMap<Integer, HashSet<Integer>> to_routes = new HashMap<>();
    for (int i = 0; i < routes.length; ++i) {
        for (int j : routes[i]) {
            if (!to_routes.containsKey(j))
                to_routes.put(j, new HashSet<Integer>());
            to_routes.get(j).add(i);
        }
    }
    Queue<int[]> bfs = new ArrayDeque();
    bfs.offer(new int[] {S, 0});
    HashSet<Integer> seen = new HashSet<>();
    seen.add(S);
    boolean[] seen_routes = new boolean[n];
    while (!bfs.isEmpty()) {
        int stop = bfs.peek()[0], bus = bfs.peek()[1];
        bfs.poll();
        if (stop == T)         return bus;
        for (int i : to_routes.get(stop)) {
            if (seen_routes[i]) continue;
            for (int j : routes[i]) {
                if (!seen.contains(j)) {
                    seen.add(j);
                    bfs.offer(new int[] {j, bus + 1});
                }
            }
            seen_routes[i] = true;
        }
    }
    return -1;
}
```


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