H
1534208810
tags: BFS
```
/*
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.
*/
/*
1. Map stop -> bus list
2. Add stop to queue
3. process queue, if stop match return; otherwise, try the linked bus (track visited bus)
*/
class Solution {
public int numBusesToDestination(int[][] routes, int S, int T) {
Set visited = new HashSet<>();
Map> stopMap = new HashMap<>();
Queue queue = new LinkedList<>();
// init
for (int i = 0; i < routes.length; i++) {
for (int stop : routes[i]) {
stopMap.putIfAbsent(stop, new HashSet<>());
stopMap.get(stop).add(i); // add bus route to stop
}
}
queue.offer(S);
int count = 0;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
int stop = queue.poll(); // 1, 2, 7
if (stop == T) return count;
for (int bus : stopMap.get(stop)) {
if (!visited.contains(bus)) {
visited.add(bus);
for (int nextStop : routes[bus]) {
queue.offer(nextStop); // 3, 6, 7
}
}
}
}
count++;
}
return -1;
}
}
```