Version #1 Classical BFS
key is how to get next buses
-> We can get the next bus from the current bus if any stop in route[curr] also contains that next bus
65.24%
public int getMinTransferNumber(int N, int[][] route, int A, int B) {
// Write your code here
Map<Integer, Set<Integer>> map = new HashMap<>();
for (int i = 0; i < N; i++) {
for (int r = 0; r < route[i].length; r++) {
map.putIfAbsent(route[i][r], new HashSet<>());
map.get(route[i][r]).add(i);
}
}
Queue<Integer> que = new LinkedList<>();
for (int initBus : map.get(A)) {
que.offer(initBus);
}
map.remove(A);
int step = 0;
while (!que.isEmpty() && !map.isEmpty()) {
step++;
int size = que.size();
while (size-- > 0) {
int currBus = que.poll();
for (int stop : route[currBus]) { // stops that can be reached from
if (stop == B) {
return step;
}
if (map.containsKey(stop)) {
Set<Integer> buses = map.get(stop);
for (int bus : buses) {
que.offer(bus);
}
map.remove(stop);
}
}
}
}
return -1;
}
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