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; } }
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