bfs

controller/algorithm/bus-routes.go

@@ -0,0 +1,167 @@
+package algorithm
+
+import "fmt"
+
+/*
+原题：https://leetcode-cn.com/problems/bus-routes/
+815. 公交路线
+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。
+
+[[7,12],[4,5,15],[6],[15,19],[9,12,13]]
+15
+12
+
+-1
+
+提示：
+
+1 <= routes.length <= 500.
+1 <= routes[i].length <= 10^5.
+0 <= routes[i][j] < 10 ^ 6.
+
+分析：
+1. 给公交车编号，找出到达开始位置的公交车和达到终点位置的公交车
+2. 使用二维数组，记录有相交车站的公交车
+3. 使用广度优先遍历，查看有相交的公交车是否能够到达终点站
+*/
+
+var minR int = -1
+
+func hasEdge(aRoute []int, bRoute []int) bool {
+	for i := 0; i < len(aRoute); i++ {
+		for j := 0; j < len(bRoute); j++ {
+			if aRoute[i] == bRoute[j] {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+func NumBusesToDestination(routes [][]int, S int, T int) int {
+	minR = -1
+	length := len(routes)
+	if length == 0 {
+		return -1
+	}
+	if S == T {
+		return 0
+	}
+	//sb和eb存储经过开始和终点的公交车编号
+	var sb, eb []int
+	for i := 0; i < length; i++ {
+		for j := 0; j < len(routes[i]); j++ {
+			if routes[i][j] == S {
+				sb = append(sb, i)
+			}
+			if routes[i][j] == T {
+				eb = append(eb, i)
+			}
+		}
+	}
+	//判断两个公交车是否有相交的车站
+	connect := make([][]int, length)
+	for i := 0; i < length-1; i++ {
+		for j := i + 1; j < length; j++ {
+			if hasEdge(routes[i], routes[j]) {
+				connect[i] = append(connect[i], j)
+				connect[j] = append(connect[j], i)
+			}
+		}
+	}
+	//fmt.Println(sb,eb,connect)
+	//方案1：深度优先遍历.从有起始车站的公交车开始遍历
+	/*for i := 0; i < len(sb); i++ {
+		r := make([]int, length)
+		r[sb[i]] = 1
+		dfsRoute(connect, sb[i], eb, r, 1)
+		r[sb[i]] = 0
+	}*/
+
+	//方案2：广度优先遍历
+	for i := 0; i < len(sb); i++ {
+		q := make([]int, 0)
+		q = append(q, sb[i])
+		bfsRoute(connect, q, eb, length)
+	}
+	fmt.Println(minR)
+	return minR
+}
+
+func inArray(data int, arr []int) bool {
+	for i := 0; i < len(arr); i++ {
+		if data == arr[i] {
+			return true
+		}
+	}
+	return false
+}
+
+//广度优先搜索
+func bfsRoute(connect [][]int, q []int, endBusList []int, length int) {
+	if inArray(q[0], endBusList) {
+		minR = 1
+		return
+	}
+	record := make([]int, length)
+	record[q[0]] = 1
+	for i := 0; i < length && i < len(q); i++ {
+		startBus := q[i]
+		for j := 0; j < len(connect[startBus]); j++ {
+			if record[connect[startBus][j]] == 0 {
+				q = append(q, connect[startBus][j])
+				record[connect[startBus][j]] = record[startBus] + 1
+				if inArray(connect[startBus][j], endBusList) {
+					if minR == -1 || minR > record[startBus]+1 {
+						minR = record[startBus] + 1
+					}
+					return
+				}
+			}
+		}
+	}
+}
+
+//深度优先遍历
+func dfsRoute(connect [][]int, startBus int, endBusList []int, r []int, l int) {
+	if minR != -1 && l >= minR {
+		return
+	}
+	if inArray(startBus, endBusList) {
+		if minR == -1 || minR > l {
+			minR = l
+		}
+		return
+	}
+	for i := 0; i < len(connect[startBus]); i++ {
+		busIndex := connect[startBus][i]
+		//fmt.Println(busIndex,r,endBusList)
+		if r[busIndex] == 0 {
+			if inArray(busIndex, endBusList) {
+				if minR == -1 || minR > l+1 {
+					minR = l + 1
+				}
+			} else {
+				r[busIndex] = l + 1
+				dfsRoute(connect, busIndex, endBusList, r, l+1)
+				r[busIndex] = 0
+			}
+		}
+	}
+}

main.go

@@ -10,8 +10,10 @@ import (
 
 func main() {
 	//nums := [][]string{{"MUC","LHR"},{"JFK","MUC"},{"SFO","SJC"},{"LHR","SFO"}}
-	nums := [][]string{{"JFK","SFO"},{"JFK","ATL"},{"SFO","ATL"},{"ATL","JFK"},{"ATL","SFO"}}
-	algorithm.FindItinerary(nums)
+	//nums := [][]int{{1, 2, 7}, {3, 6, 7}}
+	//nums := [][]int{{7,12},{4,5,15},{6},{15,19},{9,12,13}}
+	nums := [][]int{{2, 8}, {2}}
+	algorithm.NumBusesToDestination(nums,8,2)
 	r := gin.Default()
 	InitRedis()
 	router.InitRouter(r)