😫815 Bus Routes
class Solution {
public int numBusesToDestination(int[][] routes, int source, int target) {
// stop -> buses that go through this stop
Map<Integer, Set<Integer>> stopBuses = new HashMap<>();
for (int i = 0; i < routes.length; i++) {
for (int stop : routes[i]) {
Set<Integer> buses = stopBuses.get(stop);
if (buses == null) {
buses = new HashSet<>();
}
buses.add(i);
stopBuses.put(stop, buses);
}
}
int len = 0;
Queue<Integer> queue = new ArrayDeque<>();
Set<Integer> visited = new HashSet<>();
queue.offer(source);
while (!queue.isEmpty()) {
int size = queue.size();
for (int s = 0; s < size; s++) {
int stop = queue.poll();
if (stop == target) {
return len;
}
// if (visited.contains(stop)) {
// continue;
// }
// visited.add(stop);
Set<Integer> buses = stopBuses.get(stop);
for (int bus : buses) {
if (visited.contains(bus)) {
continue;
}
//???
visited.add(bus);
int[] route = routes[bus];
for (int i = 0; i < route.length; i++) {
// if (visited.contains(route[i])) {
// continue;
// }
if (route[i] == target) {
return len + 1;
}
queue.offer(route[i]);
}
}
}
len++;
}
return -1;
}
}好奇怪的题,感觉大概方向是对的,但是去重方式不对会导致tle,这里似乎是允许stop重复,但是不允许bus重复,也就是你可以去某个stop再回来,但是你不能在两次坐一样的bus,我感觉是如果两次坐一样的bus会死循环 以下是参考的解法
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