Description
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.
Discuss
使用广度遍历。首先把每一站的可达bus数存起来,也就是把每一站出现的行存起来。然后从起始站开始,把可达的每一个bus站都存入队列,依次遍历,一直到最后的终点站结束。
Code
class Solution {
public int numBusesToDestination(int[][] routes, int S, int T) {
HashSet<Integer> visited = new HashSet<>();
Queue<Integer> q = new LinkedList<>();
Map<Integer, ArrayList<Integer>> map = new HashMap<>();
int res = 0;
if (S == T) { return 0; }
for (int i = 0; i < routes.length; i++) {
for (int j = 0; j < routes[0].length; j++) {
//把每一站出现的行记录下来,如{[1,2,7],[3,6,7]}中,7出现了两次,就把0,1添加进list中
ArrayList<Integer> list = map.getOrDefault(routes[i][j], new ArrayList<Integer>());
list.add(i);
map.put(routes[i][j], list);
}
}
q.offer(S);
while (!q.isEmpty()) {
//所有与S相连的元素的个数
int len = q.size();
res++;
for (int i = 0; i < len; i++) {
int s = q.poll();
ArrayList<Integer> list = map.get(s);
for (int bus : list) {
if (visited.contains(bus)) continue;
visited.add(bus);
for (int j = 0; j < routes[bus].length; j++) {
if (routes[bus][j] == T) return res;
q.offer(routes[bus][j]);
}
}
}
}
return -1;
}
}