Leetcode 815. Bus Routes

Tyan

发布于 2021-07-08 16:26:32

38200

代码可运行

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代码可运行

文章作者：Tyan 博客：noahsnail.com | CSDN | 简书

1. Description

2. Solution

**解析：**Version 1，从起始站出发，每一次乘车，按照广度优先进行搜索所有可能到达的车站，将所有可能的车站作为候选的下一次乘车的出发站，重新进行搜索，每一次搜索过的公交车路线要从总路线中剔除，直至没有候选的乘车站为止，由于搜索了很多不能换乘的无用车站，因此超时。Version 2，在Version 1的基础上进行改进，首先遍历所有路线，只保留可以换乘的车站、起始站和终点站，然后再执行Version 1的广度优先搜索，搜索时间大幅缩短，可以超过99%的方法。

Version 1

class Solution:
    def numBusesToDestination(self, routes: List[List[int]], source: int, target: int) -> int:
        candidates = set([source])
        count = 0
        while candidates:
            if target in candidates:
                return count
            count += 1
            temp = set()
            for cand in candidates:
                indices = []
                for index, route in enumerate(routes):
                    if cand in route:
                        temp |= set(route)
                        indices.append(index)

                for i in indices[::-1]:
                    routes.pop(i)

            candidates = temp
        return -1

Version 2

class Solution:
    def numBusesToDestination(self, routes: List[List[int]], source: int, target: int) -> int:
        # stat = {}
        # for route in routes:
        #     for stop in route:
        #         stat[stop] = stat.get(stop, 0) + 1

        stat = collections.Counter([x for route in routes for x in route])

        transfers = []
        for route in routes:
            temp = []
            for stop in route:
                if stat[stop] > 1 or stop == source or stop == target:
                    temp.append(stop)
            if len(temp) > 0:
                transfers.append(temp)

        candidates = set([source])
        count = 0
        while candidates:
            if target in candidates:
                return count
            count += 1
            temp = set()
            for cand in candidates:
                indices = []
                for index, route in enumerate(transfers):
                    if cand in route:
                        temp |= set(route)
                        indices.append(index)

                for i in indices[::-1]:
                    transfers.pop(i)

            candidates = temp
        return -1