题目
https://www.lintcode.com/problem/1002
给定一个巴士路线列表 routes. routes[i] 是第 i 辆巴士的循环路线. 例如, 如果 routes[0] = [1, 5, 7], 那么第一辆巴士按照 1 -> 5 -> 7 -> 1 -> 5 -> 7 ... 的路径不停歇地行进.给定 S 和 T, 问在仅仅乘巴士的情况下, 从 S 赶到 T 最少乘多少辆不同的巴士? 如果无法赶到, 返回 -1.1 <= routes.length <= 500
1 <= routes[i].length <= 500
0 <= routes[i][j] < 10 ^ 6
样例
样例 1:输入: routes = [[1, 2, 7], [3, 6, 7]], S = 1, T = 6
输出: 2
解释: 坐第一辆车到 7, 然后坐第二辆车到 6.
样例 2:输入: routes = [[1], [15, 16, 18], [3, 4,12,14]], S = 3, T = 15
输出: -1
解释: 没有从 3 到 15 的路线.
思路
bfs
答案
public class Solution {/*** @param routes: a list of bus routes* @param s: start* @param t: destination* @return: the least number of buses we must take to reach destination*/public int numBusesToDestination(int[][] routes, int s, int t) {//和825题差不多Map<Integer, List<Integer>> stopmap = new HashMap<>();Map<Integer, List<Integer>> carmap = new HashMap<>();for (int i = 0; i <routes.length ; i++) {for (int j = 0; j < routes[i].length; j++) {int stop = routes[i][j]; //车站int car = i; //第i辆车if(!stopmap.containsKey(stop))stopmap.put(stop,new ArrayList<>());if(!carmap.containsKey(car))carmap.put(car,new ArrayList<>());stopmap.get(stop).add(car);carmap.get(car).add(stop);}}Queue<Integer> q = new LinkedList<>();Set<Integer> visited = new HashSet<>();q.add(s);visited.add(s);int steps = 0;while (!q.isEmpty()){steps++;int size = q.size();for (int i = 0; i <size ; i++) {int stop = q.poll();if(stop == t)return steps-1;for (int sp : stopmap.get(stop)) {for (int c : carmap.get(sp)) {if(visited.contains(c)) continue;q.add(c);visited.add(c);}}}}return -1;}
}
