[Leetcode][Python]36: Valid Sudoku
# -*- coding: utf8 -*-
'''
__author__ = 'dabay.wang@gmail.com' 36: Valid Sudoku
https://oj.leetcode.com/problems/valid-sudoku/ Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.
The Sudoku board could be partially filled, where empty cells are filled with the character '.'. Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated. ===Comments by Dabay===
先检查每个3x3的格子。
然后遍历board,当遇到数字时,检查其所在的行和列。这里用d_row和d_col记录检查过的行和列,避免重复检查。
''' class Solution:
# @param board, a 9x9 2D array
# @return a boolean
def isValidSudoku(self, board):
for i in xrange(0, 9, 3):
for j in xrange(0, 9, 3):
d = {}
for x in xrange(i, i+3):
for y in xrange(j, j+3):
if board[x][y] == '.':
continue
n = board[x][y]
if n in d:
return False
d[n] = True d_row = {}
d_col = {}
for i in xrange(0, 9):
for j in xrange(0, 9):
if board[i][j] == '.':
continue
num = board[i][j]
if i not in d_row:
d = {}
for x in xrange(0, 9):
if board[i][x] == '.':
continue
n = board[i][x]
if n in d:
return False
d[n] = True
d_row[i] = True
if j not in d_col:
d = {}
for y in xrange(0, 9):
if board[y][j] == '.':
continue
n = board[y][j]
if n in d:
return False
d[n] = True
d_col[j] = True
return True def main():
sol = Solution()
board = [
"53..7....",
"6..195...",
".98....6.",
"8...6...3",
"4..8.3..1",
"7...2...6",
".6....28.",
"...419..5",
"....8..79"
]
print sol.isValidSudoku(board) if __name__ == '__main__':
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)
[Leetcode][Python]36: Valid Sudoku的更多相关文章
- 【LeetCode】36. Valid Sudoku 解题报告(Python)
[LeetCode]36. Valid Sudoku 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址 ...
- 【LeetCode】36 - Valid Sudoku
Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.(http://sudoku.com.au/TheRu ...
- 【一天一道LeetCode】#36. Valid Sudoku
一天一道LeetCode 本系列文章已全部上传至我的github,地址:https://github.com/Zeecoders/LeetCode 欢迎转载,转载请注明出处 (一)题目 Determi ...
- 【leetcode】36. Valid Sudoku(判断能否是合法的数独puzzle)
Share Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated accordi ...
- LeetCode:36. Valid Sudoku(Medium)
1. 原题链接 https://leetcode.com/problems/valid-sudoku/description/ 2. 题目要求 给定一个 9✖️9 的数独,判断该数独是否合法 数独用字 ...
- LeetCode:36. Valid Sudoku,数独是否有效
LeetCode:36. Valid Sudoku,数独是否有效 : 题目: LeetCode:36. Valid Sudoku 描述: Determine if a Sudoku is valid, ...
- leetcode 37. Sudoku Solver 36. Valid Sudoku 数独问题
三星机试也考了类似的题目,只不过是要针对给出的数独修改其中三个错误数字,总过10个测试用例只过了3个与世界500强无缘了 36. Valid Sudoku Determine if a Sudoku ...
- [LeetCode] 36. Valid Sudoku 验证数独
Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to th ...
- 蜗牛慢慢爬 LeetCode 36.Valid Sudoku [Difficulty: Medium]
题目 Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board could ...
随机推荐
- CSS自学笔记(13):CSS3 2D/3D转换
CSS3中新增了对元素进行2D和3D的转换效果,这样可以是开发人员很方便的做出视觉效果更好的网页来. 通过CSS3中属性的定义,我们可以对元素进行移动.缩放.拉伸.旋转等等,可以通过定义transfo ...
- 数据结构-B树
1.前言: 动态查找树主要有:二叉查找树(Binary Search Tree),平衡二叉查找树(Balanced Binary Search Tree),红黑树(Red-Black Tree ) ...
- 大型网站性能优化(页面(HTML)优化的方法)
页面(HTML)优化的方法 除了语言层面上进行优化外,对Web开发,HTML的优化将很大程度上减轻服务器的负载,提高网站的性能 1). 减少HTTP请求数.打开网页,浏览器会发出很多请求,图片,脚本, ...
- Keil MDK中使用pc-lint的详细方法
keil MDK版本:V4.03 PC-lint版本: V8.0 关于pc-lint的强大作用,网上有很多,这里不想再复述,只说一句:能通过pc-lint检验的程序不一定没有问题,但通过了pc-li ...
- Linux进程间通信——使用信号量
这篇文章将讲述别一种进程间通信的机制——信号量.注意请不要把它与之前所说的信号混淆起来,信号与信号量是不同的两种事物.有关信号的更多内容,可以阅读我的另一篇文章:Linux进程间通信——使用信号.下面 ...
- windows、linux创建子进程
在windows下创建子进程较常用到的API就是CreateProcess,可以通过以下的方式启动一个新进程: STARTUPINFO si = {0}; PROCES ...
- poj3094---对字符串的处理
#include <stdio.h> #include <stdlib.h> #include<string.h> int main() { ]; int len, ...
- Spring、Spring自动扫描和管理Bean
Spring2.5为我们引入了组件自动扫描机制,它可以在类路径下寻找标记了@Component.@Service.@Controller.@Repository注解的类,并把这些类纳入到spring容 ...
- Matrix(类似kruskal)
Matrix Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Submis ...
- python代码中pass的用法
我们有时会在方法中写一些注释代码,用来提示这个方法是干嘛的之类,看下面代码: class Game_object: def __init__(self, name): self.name = name ...