😫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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