标签:inpu def elf 解题思路 obj ber tde bisect span
题目如下:
You are given the number of rows
n_rows
and number of columnsn_cols
of a 2D binary matrix where all values are initially 0. Write a functionflip
which chooses a 0 value uniformly at random, changes it to 1, and then returns the position[row.id, col.id]
of that value. Also, write a functionreset
which sets all values back to 0. Try to minimize the number of calls to system‘s Math.random() and optimize the time and space complexity.Note:
1 <= n_rows, n_cols <= 10000
0 <= row.id < n_rows
and0 <= col.id < n_cols
flip
will not be called when the matrix has no 0 values left.- the total number of calls to
flip
andreset
will not exceed 1000.Example 1:
Input: ["Solution","flip","flip","flip","flip"] [[2,3],[],[],[],[]] Output: [null,[0,1],[1,2],[1,0],[1,1]]
Example 2:
Input: ["Solution","flip","flip","reset","flip"] [[1,2],[],[],[],[]] Output: [null,[0,0],[0,1],null,[0,0]]
Explanation of Input Syntax:
The input is two lists: the subroutines called and their arguments.
Solution
‘s constructor has two arguments,n_rows
andn_cols
.flip
andreset
have no arguments. Arguments are always wrapped with a list, even if there aren‘t any.
解题思路:每一个row最多可以生成col次,我的想法是创建一个row_pool,同时有一个字典保存每个row已经使用的次数。如果已经使用了col次,则把row从对应的row_pool里面删除,这样就可以保证只用一次随机就可以在row_pool里面找到可用的row。col也是一样的原理,用第二个字典记录每个row还能使用的col列表,每使用一个col,就从col列表中删除,保证只用一次随机就可以在col列表里面找到可用的col。
代码如下:
class Solution(object): def __init__(self, n_rows, n_cols): """ :type n_rows: int :type n_cols: int """ self.rows_pool = range(n_rows) self.dic_rows_count = {} self.row_can_use_cols = {} self.row = n_rows self.col = n_cols def flip(self): """ :rtype: List[int] """ import random import bisect r = random.randint(0, len(self.rows_pool)-1) r = self.rows_pool[r] self.dic_rows_count[r] = self.dic_rows_count.setdefault(r,0) + 1 if self.dic_rows_count[r] == self.col: del self.rows_pool[bisect.bisect_left(self.rows_pool,r)] if r not in self.row_can_use_cols: self.row_can_use_cols[r] = range(self.col) c = random.randint(0, len(self.row_can_use_cols[r]) - 1) c = self.row_can_use_cols[r][c] del self.row_can_use_cols[r][bisect.bisect_left(self.row_can_use_cols[r], c)] return [r,c] def reset(self): """ :rtype: None """ self.rows_pool = range(self.row ) self.dic_rows_count = {} self.row_can_use_cols = {} # Your Solution object will be instantiated and called as such: # obj = Solution(n_rows, n_cols) # param_1 = obj.flip() # obj.reset()
【leetcode】519. Random Flip Matrix
标签:inpu def elf 解题思路 obj ber tde bisect span
原文地址:https://www.cnblogs.com/seyjs/p/10489160.html