★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： https://www.cnblogs.com/strengthen/p/10564150.html
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
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。
示例: 输入: routes = [[1, 2, 7], [3, 6, 7]] S = 1 T = 6 输出: 2 解释: 最优策略是先乘坐第一辆公交车到达车站 7, 然后换乘第二辆公交车到车站 6。
说明:
1 <= routes.length <= 500.1 <= routes[i].length <= 500.0 <= routes[i][j] < 10 ^ 6.
584ms
1 class Solution { 2 // BFS is easier, also faster 3 func numBusesToDestination(_ routes: [[Int]], _ S: Int, _ T: Int) -> Int { 4 if S == T { return 0 } 5 var stopToRoutes = [Int: [Int]]() 6 for (index, route) in routes.enumerated() { 7 for stop in route { 8 if var buses = stopToRoutes[stop] { 9 buses.append(index) 10 stopToRoutes[stop] = buses 11 } else { 12 stopToRoutes[stop] = [index] 13 } 14 } 15 } 16 17 //print(stopToRoutes) 18 19 var queue = [Int]() 20 queue.append(S) 21 var visited = Set<Int>() 22 var ans = 0 23 while (!queue.isEmpty) { 24 //print(queue, visited, ans) 25 ans += 1 26 var size = queue.count 27 while (size > 0) { 28 let last = queue.remove(at: 0) 29 for route in (stopToRoutes[last] ?? []) { 30 if visited.contains(route) { 31 continue 32 } else { 33 visited.insert(route) 34 } 35 for stop in routes[route] { 36 if stop == T { 37 return ans 38 } 39 queue.append(stop) 40 } 41 } 42 size -= 1 43 } 44 } 45 return -1 46 } 47 48 // LTE 49 // graph search: DFS 50 func numBusesToDestination1(_ routes: [[Int]], _ S: Int, _ T: Int) -> Int { 51 if S == T { return 0 } 52 var stopToRoutes = [Int: [Int]]() 53 for (index, route) in routes.enumerated() { 54 for stop in route { 55 if var buses = stopToRoutes[stop] { 56 buses.append(index) 57 stopToRoutes[stop] = buses 58 } else { 59 stopToRoutes[stop] = [index] 60 } 61 } 62 } 63 print(stopToRoutes) 64 65 var startI = stopToRoutes[S] ?? [] 66 var targetI = stopToRoutes[T] ?? [] 67 var ret = Int.max 68 var routes = routes 69 for si in startI { 70 var visited = Set<Int>() 71 find(&routes, &stopToRoutes, si, targetI, &visited, 1, &ret) 72 } 73 return ret > routes.count ? -1: ret 74 } 75 76 private func find(_ routes: inout [[Int]], _ map: inout [Int: [Int]], _ curi: Int, _ ei: [Int], _ visited: inout Set<Int>, _ curStops: Int, _ ret: inout Int) { 77 if ei.contains(curi) { 78 ret = min(ret, curStops) 79 return 80 } 81 if visited.contains(curi) { 82 return 83 } 84 visited.insert(curi) 85 86 let stops = routes[curi] 87 for stop in stops { 88 let buses = map[stop] ?? [] 89 for route in buses { 90 //print("find",curi, ei, stop, route, curStops) 91 92 find(&routes, &map, route, ei, &visited, curStops + 1, &ret) 93 } 94 } 95 96 visited.remove(curi) 97 } 98 }
Runtime: 604 ms
Memory Usage: 24.5 MB
1 class Solution { 2 func numBusesToDestination(_ routes: [[Int]], _ S: Int, _ T: Int) -> Int { 3 if S == T {return 0} 4 var res:Int = 0 5 var stop2bus:[Int:[Int]] = [Int:[Int]]() 6 var q:[Int] = [S] 7 var visited:Set<Int> = Set<Int>() 8 for i in 0..<routes.count 9 { 10 for j in routes[i] 11 { 12 stop2bus[j,default:[Int]()].append(i) 13 } 14 } 15 while(!q.isEmpty) 16 { 17 res += 1 18 for i in stride(from:q.count,to:0,by:-1) 19 { 20 var t:Int = q.removeFirst() 21 for bus in stop2bus[t,default:[Int]()] 22 { 23 if visited.contains(bus) {continue} 24 visited.insert(bus) 25 for stop in routes[bus] 26 { 27 if stop == T {return res} 28 q.append(stop) 29 } 30 } 31 } 32 } 33 return -1 34 } 35 }
644ms
1 class Solution { 2 func numBusesToDestination(_ routes: [[Int]], _ S: Int, _ T: Int) -> Int { 3 var visited: Set<Int> = [] 4 var queue: [Int] = [] 5 var map: [Int: [Int]] = [:] 6 var steps = 0 7 8 if S == T { 9 return 0 10 } 11 12 for i in 0 ..< routes.count { 13 for j in 0 ..< routes[i].count { 14 if map[routes[i][j]] == nil { 15 map[routes[i][j]] = [] 16 } 17 18 map[routes[i][j]]!.append(i) 19 } 20 } 21 22 queue.append(S) 23 24 while !queue.isEmpty { 25 var size = queue.count 26 steps += 1 27 while size > 0 { 28 let curr = queue.removeFirst() 29 let buses = map[curr]! 30 for bus in buses { 31 if !visited.contains(bus) { 32 visited.insert(bus) 33 for j in 0 ..< routes[bus].count { 34 if routes[bus][j] == T { 35 return steps 36 } 37 queue.append(routes[bus][j]) 38 } 39 } 40 } 41 size -= 1 42 } 43 } 44 45 return -1 46 } 47 }
1360ms
1 class Solution { 2 func numBusesToDestination(_ routes: [[Int]], _ S: Int, _ T: Int) -> Int { 3 var stops: [Int: [Int]] = [:] 4 for (bus, route) in routes.enumerated() { 5 for stop in route { 6 stops[stop, default: []].append(bus) 7 } 8 } 9 var res = 0 10 var queue: [Int] = [] 11 queue.append(S) 12 var visitedStops: Set<Int> = [] 13 var visitedBuses: Set<Int> = [] 14 while queue.isEmpty == false { 15 var n = queue.count 16 while n > 0 { 17 n -= 1 18 let cur = queue.removeFirst() 19 if cur == T { 20 return res 21 } 22 if visitedStops.contains(cur) { 23 continue 24 } 25 visitedStops.insert(cur) 26 for bus in (stops[cur] ?? []) { 27 if visitedBuses.contains(bus) == true { 28 continue 29 } 30 visitedBuses.insert(bus) 31 for stop in routes[bus] { 32 queue.append(stop) 33 } 34 } 35 } 36 res += 1 37 } 38 return -1 39 } 40 }