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# 815.Bus-Routes

## 题目描述

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 = [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.

## 代码

### Approach 1: BFS

class Solution {
public int numBusesToDestination(int[][] routes, int S, int T) {
HashSet<Integer> visited = new HashSet<>();
HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();
int ret = 0;
if (S == T) return 0;
for (int i = 0; i < routes.length; i++) {
for (int j = 0; j < routes[i].length; j++) {
ArrayList<Integer> buses = map.getOrDefault(routes[i][j], new ArrayList<>());
map.put(routes[i][j], buses);
}
}
q.offer(S);
while (!q.isEmpty()) {
int len = q.size();
ret++;
for (int i = 0; i < len; i++) {
int cur = q.poll();
ArrayList<Integer> buses = map.get(cur);
for (int bus: buses) {
if (visited.contains(bus)) continue;
for (int j = 0; j < routes[bus].length; j++) {
if (routes[bus][j] == T) return ret;
q.offer(routes[bus][j]);
}
}
}
}
return -1;
}
}

### Approach #2 BFS

public int numBusesToDestination(int[][] routes, int S, int T) {
int n = routes.length;
HashMap<Integer, HashSet<Integer>> to_routes = new HashMap<>();
for (int i = 0; i < routes.length; ++i) {
for (int j : routes[i]) {
if (!to_routes.containsKey(j))
to_routes.put(j, new HashSet<Integer>());
}
}
Queue<int[]> bfs = new ArrayDeque();
bfs.offer(new int[] {S, 0});
HashSet<Integer> seen = new HashSet<>();
boolean[] seen_routes = new boolean[n];
while (!bfs.isEmpty()) {
int stop = bfs.peek(), bus = bfs.peek();
bfs.poll();
if (stop == T) return bus;
for (int i : to_routes.get(stop)) {
if (seen_routes[i]) continue;
for (int j : routes[i]) {
if (!seen.contains(j)) {