Create 815._Bus_Routes.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/815._Bus_Routes.md

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+# 815. Bus Routes
+
+**<font color=red>难度: Hard</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/bus-routes/description/
+
+> 内容描述
+
+```
+
+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.
+````
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(VE)******- 空间复杂度: O(VE)******
+
+看作是一个图的话，V是bus，E是bus所在的routes，那么时间和空间复杂度都是O(VE)
+
+```python
+class Solution(object):
+    def numBusesToDestination(self, routes, S, T):
+        """
+        :type routes: List[List[int]]
+        :type S: int
+        :type T: int
+        :rtype: int
+        """
+        bus_in_routes, cur_buses, step, visited = collections.defaultdict(set), [S], 0, set()
+        for i, route in enumerate(routes):
+            for bus in route:
+                bus_in_routes[bus].add(i) # this bus is in route i
+        while cur_buses:
+            new_buses = []
+            for bus in cur_buses:
+                if bus == T:
+                    return step
+                for route_idx in bus_in_routes[bus]:
+                    if route_idx not in visited:
+                        visited.add(route_idx)
+                        for new_bus in routes[route_idx]:
+                            if new_bus != bus:
+                                new_buses.append(new_bus)
+            cur_buses = new_buses
+            step += 1
+        return -1
+```