问题描述

谷歌面试题，求大神帮忙

原题是这样的（后面我补充了中文解释）：

Don't mind the map

After the trauma of Dr. Boolean's lab, the rabbits are eager to get back to their normal lives in a well-connected community, where they can visit each other frequently. Fortunately, the rabbits learned something about engineering as part of their escape from the lab. To get around their new warren fast, they built an elaborate subway system to connect their holes. Each station has the same number of outgoing subway lines (outgoing tracks), which are numbered.

Unfortunately, sections of warrens are very similar, so they can't tell where they are in the subway system. Their stations have system maps, but not an indicator showing which station the map is in. Needless to say, rabbits get lost in the subway system often. The rabbits adopted an interesting custom to account for this: Whenever they are lost, they take the subway lines in a particular order, and end up at a known station.

For example, say there were three stations A, B, and C, with two outgoing directions, and the stations were connected as follows

Line 1 from A, goes to B. Line 2 from A goes to C.
Line 1 from B, goes to A. Line 2 from B goes to C.
Line 1 from C, goes to B. Line 2 from C goes to A.

Now, suppose you are lost at one of the stations A, B, or C. Independent of where you are, if you take line 2, and then line 1, you always end up at station B. Having a path that takes everyone to the same place is called a meeting path.
We are interested in finding a meeting path which consists of a fixed set of instructions like, 'take line 1, then line 2,' etc. It is possible that you might visit a station multiple times. It is also possible that such a path might not exist. However, subway stations periodically close for maintenance. If a station is closed, then the paths that would normally go to that station, go to the next station in the same direction. As a special case, if the track still goes to the closed station after that rule, then it comes back to the originating station. Closing a station might allow for a meeting path where previously none existed. That is, if you have
A -> B -> C
and station B closes, then you'll have
A -> C
Alternately, if it was
A -> B -> B
then closing station B yields
A -> A

Write a function answer(subway) that returns one of:

-1 (minus one): If there is a meeting path without closing a station
The least index of the station to close that allows for a meeting path or
-2 (minus two): If even with closing 1 station, there is no meeting path.
subway will be a list of lists of integers such that subway[station][direction] = destination_station.

That is, the subway stations are numbered 0, 1, 2, and so on. The k^th element of subway (counting from 0) will give the list of stations directly reachable from station k.

The outgoing lines are numbered 0, 1, 2... The r^th element of the list for station k, gives the number of the station directly reachable by taking line r from station k.

Each element of subway will have the same number of elements (so, each station has the same number of outgoing lines), which will be between 1 and 5.

There will be at least 1 and no more than 50 stations.

For example, if
subway = [[2, 1], [2, 0], [3, 1], [1, 0]]
Then one could take the path [1, 0]. That is, from the starting station, take the second direction, then the first. If the first direction was the red line, and the second was the green line, you could phrase this as:
if you are lost, take the green line for 1 stop, then the red line for 1 stop.
So, consider following the directions starting at each station.
0 -> 1 -> 2.
1 -> 0 -> 2.
2 -> 1 -> 2.
3 -> 0 -> 2.
So, no matter the starting station, the path leads to station 2. Thus, for this subway, answer should return -1.

If
subway = [[1], [0]]
then no matter what path you take, you will always be at a different station than if you started elsewhere. If station 0 closed, that would leave you with
subway = [[0]]
So, in this case, answer would return 0 because there is no meeting path until you close station 0.

To illustrate closing stations,
subway = [[1,1],[2,2],[0,2]]
If station 2 is closed, then
station 1 direction 0 will follow station 2 direction 0 to station 0, which will then be its new destination.
station 1 direction 1 will follow station 2 direction 1 to station 2, but that station is closed, so it will get routed back to station 1, which will be its new destination. This yields
subway = [[1,1],[0,1]]

Languages

To provide a Python solution, edit solution.py
To provide a Java solution, edit solution.java

Test cases

Inputs:
(int) subway = [[2, 1], [2, 0], [3, 1], [1, 0]]
Output:
(int) -1

Inputs:
(int) subway = [[1, 2], [1, 1], [2, 2]]
Output:
(int) 1

大概意思就是，兔子在兔子窝(n <= 50)之间修了地铁，有k(k <= 5)条线路，每条线每一站都有出发，但是不一定每站都到，比如1号线可能是A->B, B->A, C->B; 2号线可能是A->C, B->C, C->A. 由于兔子窝长得都一样，所以兔子经常不知道自己究竟在哪一站，所以它们想了个办法，在任何一站，只要按照某种顺序乘坐地铁，最终都会到达某一站，比如上面的线路，只要按照先坐一号线，再坐二号线的顺序，无论在ABC哪一站出发，最后都会到C站。现在给出一种地铁线路，问存不存在这样一种顺序，满足按照这个顺序坐地铁，从任何一站出发都能到达某一站。以及如果有一站地铁关门维修了（如果某站关门了，那么到达这一站的车这一站不停，直接前往下一站，如果这一站的下一站还是自己，那么就在上一站循环），是否还存在这样一种顺序。

想了两天了，暴力解的话2^50严重超时，跪求大神帮忙

解决方案

solved, thanks