Leetcode 3243题解

class Solution:
    def shortestDistanceAfterQueries(self, n: int, queries):
        neighbors = [[i + 1] for i in range(n)]
        neighbors[-1] = []
        res = []
        for [u, v] in queries:
            neighbors[u].append(v)
            res.append(self.bfs(n, neighbors))

        return res

    def bfs(self, n: int, neighbors):
        from collections import deque
        #记录从0到N的最短路径
        dist = [-1] * n
        dist[0] = 0
        #指针，记录当前要赋的x值，即查找值
        q = deque([0])
        while (len(q)) > 0:
            #横坐标x值
            x = q.popleft()
            #纵坐标y值
            for y in neighbors[x]:
                if dist[y] >= 0:
                    continue
                q.append(y)
                #X和Y建立连接
                dist[y] = dist[x] + 1

        return dist[n - 1]


if __name__ == '__main__':
    n = 5
    queries = [[2, 4], [0, 2], [0, 4]]
    solution = Solution()
    res = solution.shortestDistanceAfterQueries(n, queries)
    print("##########res:##########", res)