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添加 problem 815
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38
Algorithms/0815. Bus Routes/815. Bus Routes.go
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38
Algorithms/0815. Bus Routes/815. Bus Routes.go
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package leetcode
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func numBusesToDestination(routes [][]int, S int, T int) int {
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if S == T {
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return 0
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}
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// 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
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vertexMap, visited, queue, res := map[int][]int{}, make([]bool, len(routes)), []int{}, 0
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for i := 0; i < len(routes); i++ {
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for _, v := range routes[i] {
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tmp := vertexMap[v]
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tmp = append(tmp, i)
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vertexMap[v] = tmp
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}
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}
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queue = append(queue, S)
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for len(queue) > 0 {
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res++
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qlen := len(queue)
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for i := 0; i < qlen; i++ {
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vertex := queue[0]
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queue = queue[1:]
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for _, bus := range vertexMap[vertex] {
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if visited[bus] == true {
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continue
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}
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visited[bus] = true
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for _, v := range routes[bus] {
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if v == T {
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return res
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}
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queue = append(queue, v)
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}
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}
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}
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}
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return -1
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}
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44
Algorithms/0815. Bus Routes/815. Bus Routes_test.go
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44
Algorithms/0815. Bus Routes/815. Bus Routes_test.go
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question815 struct {
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para815
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ans815
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}
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// para 是参数
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// one 代表第一个参数
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type para815 struct {
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r [][]int
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s int
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t int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans815 struct {
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one int
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}
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func Test_Problem815(t *testing.T) {
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qs := []question815{
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question815{
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para815{[][]int{[]int{1, 2, 7}, []int{3, 6, 7}}, 1, 6},
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ans815{2},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 815------------------------\n")
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for _, q := range qs {
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_, p := q.ans815, q.para815
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fmt.Printf("【input】:%v 【output】:%v\n", p, numBusesToDestination(p.r, p.s, p.t))
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}
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fmt.Printf("\n\n\n")
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}
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41
Algorithms/0815. Bus Routes/README.md
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41
Algorithms/0815. Bus Routes/README.md
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# [815. Bus Routes](https://leetcode.com/problems/bus-routes/)
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## 题目:
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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.
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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.
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Example:
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Input:
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routes = [[1, 2, 7], [3, 6, 7]]
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S = 1
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T = 6
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Output: 2
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Explanation:
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The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.
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**Note:**
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- `1 <= routes.length <= 500`.
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- `1 <= routes[i].length <= 500`.
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- `0 <= routes[i][j] < 10 ^ 6`.
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## 题目大意
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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。
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说明:
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- 1 <= routes.length <= 500.
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- 1 <= routes[i].length <= 500.
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- 0 <= routes[i][j] < 10 ^ 6.
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## 解题思路
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- 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?
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- 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。
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@@ -879,7 +879,7 @@
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| 0812 | Largest Triangle Area | | 56.30% | Easy | |
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| 0813 | Largest Sum of Averages | | 45.50% | Medium | |
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| 0814 | Binary Tree Pruning | | 71.20% | Medium | |
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| 0815 | Bus Routes | | 40.30% | Hard | |
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| 0815 | Bus Routes | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0815.%20Bus%20Routes) | 40.30% | Hard | |
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| 0816 | Ambiguous Coordinates | | 44.20% | Medium | |
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| 0817 | Linked List Components | [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0817.%20Linked%20List%20Components) | 54.80% | Medium | |
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| 0818 | Race Car | | 35.20% | Hard | |
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