815. Bus Routes
Posted yaoyudadudu
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了815. Bus Routes相关的知识,希望对你有一定的参考价值。
问题描述:
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
.
解题思路:
此时我们有了车到站的对应,我们可以建立一个站到车的对应。
用hash表来辅助我们建立:unordered_map<int, vector<int>> m;
这时候我们其实可以将这些路线的叠加看成一个图。
我们从起点开始bfs遍历:将起点压入队列中,进行层序遍历。
由于我们要求的事最少的换车数,这代表我们同一辆车只经过一次,同一个站点也只经过一次,可以用集合来存储VisitedStops以及VisitedBus。
我们可以用队列的大小size来帮助我们进行层序遍历。
代码:
class Solution { public: int numBusesToDestination(vector<vector<int>>& routes, int S, int T) { unordered_map<int, vector<int>> m; for(int i = 0 ; i < routes.size(); i++){ for(int n : routes[i]) m[n].push_back(i); } int ret = 0; queue<int> q; q.push(S); unordered_set<int> visitedStops; unordered_set<int> visitedBus; while(!q.empty()){ int n = q.size(); unordered_set<int> tmp; for(int i = 0; i < n; i++){ int cur = q.front(); q.pop(); if(cur == T) return ret; if(visitedStops.count(cur)) continue; visitedStops.insert(cur); for(auto bus : m[cur]){ if(visitedBus.count(bus)) continue; visitedBus.insert(bus); for(auto stop : routes[bus]) tmp.insert(stop); } } for(auto s : tmp) q.push(s); ret++; } return -1; } };
以上是关于815. Bus Routes的主要内容,如果未能解决你的问题,请参考以下文章
[LeetCode] 815. Bus Routes 巴士路线
LeetCode 815. 公交路线 / 909. 蛇梯棋(还是bfs)/ 168. Excel表列名称 / 171. Excel表列序号