for row in range(9): c = board[row][col] if c == '.': continue if c in exist_set: return False else: exist_set.add(c) # - check for each grid for gr in range(3): for gc in range(3): exist_set = set() for j in range(3): for i in range(3): c = board[gr*3...
Do not return anything, modify board in-place instead. """row_list = [[False]*10for_inrange(9)] col_list = [[False]*10for_inrange(9)] grid_list = [[False]*10for_inrange(9)]# - get grid indexdefget_grid_index(row,col):return(row//3)*3+ col//3# - set state to value...
$ python csv_to_requirements.py royle.csv - | uv pip compile --find-links packages/ --no-annotate --no-header - | python render_solution.py - Resolved 81 packages in 126ms 5,3,4,6,7,8,9,1,2 6,7,2,1,9,5,3,4,8 1,9,8,3,4,2,5,6,7 8,5,9,7,6,1,4,2,3 4,2...
seen or (c, y) in self.seen or (x/3, y/3, c) in self.seen: print 'buxing' return False self.seen.add((x, c)) self.seen.add((c, y)) self.seen.add((x/3, y/3, c)) return True def dfs(self,board): for i in range(9): for j in range(9): if board[i][j] =...
【Sudoku in Python Packaging:用Python包管理器解数独,将数独游戏转化为依赖版本解析问题,以创新方式解决传统逻辑游戏】'sudoku-in-python-packaging - Solve sudokus not in python, but in python packages.' GitHub: github.com/konstin/sudoku-in-python-packaging #数独# #Python# #包管理器# ...
def validSolution(board): blocks = [[board[x+a][y+b] for a in (0, 1, 2) for b in (0, 1, 2)] for x in (0, 3, 6) for y in (0, 3, 6)] return not filter(lambda x: set(x) != set(range(1, 10)), board + zip(*board) + blocks) 1 2 3 讲解 Python filter...
Sudoku+ Brain Teaser更多来自此开发人员的 App C++ Compiler - Run .cpp Code 教育 教育 Compiler+ - All in 1 Compiler Learn Go - OnePercent 教育 教育 Dictionary+ - Words & Phrases Learn Kotlin - OnePercent 教育 Learn CSS - OnePercent
Python gruhn/typescript-sudoku Star490 Code Issues Pull requests Playing Sudoku in TypeScript while the type checker highlights mistakes. typescripttype-theorysudoku UpdatedAug 20, 2024 TypeScript kaajjo/LibreSudoku Star468 LibreSudoku - Sudoku app for android built with Kotlin + Jetpack Compose + ...
[Leetcode][python]Sudoku Solver/解数独 题目大意 计算数独,假设解唯一 解题思路 回溯法,深度优先 代码 这一题注释写的很多,因为比较复杂头疼中 class Solution(object): seen = set() def isValue(self,board,x,y): # 判断符合,就是上一题 c = board[x][y]...
以下为译文： 时不时地，我会发现一些编程语言所做的一些与众不同的事情，也因此改变了我对编码的看法...