815. Bus Routes #

题目 #

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.

题目大意 #

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:

1 <= routes.length <= 500.
1 <= routes[i].length <= 500.
0 <= routes[i][j] < 10 ^ 6.

解题思路 #

给出一些公交路线，公交路径代表经过的哪些站。现在给出起点和终点站，问最少需要换多少辆公交车才能从起点到终点？
这一题可以转换成图论的问题，将每个站台看成顶点，公交路径看成每个顶点的边。同一个公交的边染色相同。题目即可转化为从顶点 S 到顶点 T 需要经过最少多少条不同的染色边。用 BFS 即可轻松解决。从起点 S 开始，不断的扩展它能到达的站点。用 visited 数组防止放入已经可达的站点引起的环。用 map 存储站点和公交车的映射关系(即某个站点可以由哪些公交车到达)，BFS 的过程中可以用这个映射关系，拿到公交车的其他站点信息，从而扩张队列里面的可达站点。一旦扩展出现了终点 T，就可以返回结果了。

代码 #


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
}

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