815. Bus Routes

815. Bus Routes

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.

------------------------
Solution #1 以bus route为⼀个node, 构建graph O(n * n * l) time, n为route的数量量，l为route的平均长度

class Solution {
    public int numBusesToDestination(int[][] routes, int s, int t) {
        if (s == t) return 0;
        
        Map<Integer, Cluster> graph = new HashMap<>();
        Queue<Integer> source = mergeRoutes(graph, routes, s);
        Set<Integer> visited = new HashSet<>();
        int len = 1;
        
        while (!source.isEmpty()) {
            
            int size = source.size();
            while (size > 0) {
                int top = source.poll();
                
                size--;
                if (visited.contains(top)) continue;
                visited.add(top);
                Cluster topC = graph.get(top);
                
                if (topC.elems.contains(t)) return len;
                for (int c : topC.neighbours) {
                    source.add(c);
                }
            }
            len++;
        }
        
        return -1;
    }
    
    private Queue<Integer> mergeRoutes(Map<Integer, Cluster> graph, int[][] routes, int s) {
        
        Queue<Integer> source = new LinkedList<>();
        
        for (int i = 0; i < routes.length; i++) {
            Cluster c = new Cluster(i, routes[i]);
            graph.put(i, c);
            
            for (int elem : routes[i]) {
                if (elem == s) source.add(i);
                    
                for (Map.Entry<Integer, Cluster> entry : graph.entrySet()) {
                    if (entry.getValue().elems.contains(elem)) {
                        entry.getValue().neighbours.add(i);
                        c.neighbours.add(entry.getValue().index);
                    }
                }
            }
        }

        return source;
    }
    
    class Cluster{
        public int index;
        public Set<Integer> elems;
        public Set<Integer> neighbours;
        
        public Cluster(int index, int[] route) {
            this.index = index;
            elems = new HashSet<>();
            for (int i : route) {
                elems.add(i);
            }
            
            neighbours = new HashSet<>();
        }
    }
}

Solution#2, 可以用stop当作node构建graph，与上面是相同复杂度
ref: https://leetcode.com/problems/bus-routes/discuss/122712/Simple-Java-Solution-using-BFS?page=1
Node是stop，neighbour是bus routes，visited里面存的也是bus route

class Solution {
    public int numBusesToDestination(int[][] routes, int s, int t) {
        if (s == t) return 0;
        
        Map<Integer, List<Integer>> stopToBus = new HashMap<>();
        Set<Integer> visited = new HashSet<>();
        
        for (int i = 0; i < routes.length; i++) {
            for (int j = 0; j < routes[i].length; j++) {
                int stop = routes[i][j];
                List<Integer> ro = stopToBus.getOrDefault(stop, new ArrayList<Integer>());
                ro.add(i);
                stopToBus.put(stop, ro);
            }
        }
        
        int r = 0;
        Queue<Integer> que = new LinkedList<>();
        que.add(s);
        
        while (!que.isEmpty()) {
            int len = que.size();
            r++;
            for (int i = 0; i < len; i++) {
                int cur = que.poll();    
                for (int bus : stopToBus.get(cur)) {
                    if (visited.contains(bus)) continue;
                    visited.add(bus);
                    for (int stop : routes[bus]) {
                        if (stop == t) return r;
                        que.add(stop);
                    }
                }
            }
        }
        
        return -1;
    }
}