Google Interview - 警察到房间的最短距离 - 我的博客 - ITeye技术网站



Google Interview - 警察到房间的最短距离 - 我的博客 - ITeye技术网站

一个 n x n 矩阵,每个房间可能是封闭的房间,可能是警察,可能是开的房间, 封闭的房间不能过,返回一个n x n矩阵,每一个元素是最近的警察到这个房间的最短距离。 初始矩阵中-1代表封闭房间,INT_MAX代表普通房间,0代表有警察的房间。

 

Solution: 把警察都找出来,然后一起push到BFS的queue里面,同时搜索。复杂度可降为O(n^2)。

Starting from police and override the distance with smaller value if multiple polices can reach the same office. 


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