website/content/ChapterFour/0800~0899/0815.Bus-Routes.md
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.我们有一系列公交路线。每一条路线 routes[i] 上都有一辆公交车在上面循环行驶。例如,有一条路线 routes[0] = [1, 5, 7],表示第一辆 (下标为0) 公交车会一直按照 1->5->7->1->5->7->1->... 的车站路线行驶。假设我们从 S 车站开始(初始时不在公交车上),要去往 T 站。 期间仅可乘坐公交车,求出最少乘坐的公交车数量。返回 -1 表示不可能到达终点车站。
说明:
package leetcode
func numBusesToDestination(routes [][]int, S int, T int) int {
if S == T {
return 0
}
// vertexMap 中 key 是站点,value 是公交车数组,代表这些公交车路线可以到达此站点
vertexMap, visited, queue, res := map[int][]int{}, make([]bool, len(routes)), []int{}, 0
for i := 0; i < len(routes); i++ {
for _, v := range routes[i] {
tmp := vertexMap[v]
tmp = append(tmp, i)
vertexMap[v] = tmp
}
}
queue = append(queue, S)
for len(queue) > 0 {
res++
qlen := len(queue)
for i := 0; i < qlen; i++ {
vertex := queue[0]
queue = queue[1:]
for _, bus := range vertexMap[vertex] {
if visited[bus] == true {
continue
}
visited[bus] = true
for _, v := range routes[bus] {
if v == T {
return res
}
queue = append(queue, v)
}
}
}
}
return -1
}