数独大家应该都玩过，作为一款益智小游戏，其实是很打发时间的，但是有时候做不出来也很让人抓狂。如果有时候解谜题需要快速地进行数独求解，不妨试试使用跑一趟python代码来解决。

代码如下：

import time
t0=time.time()
class point:
	def __init__(self,x,y):
		self.x=x
		self.y=y
		self.available=[]
		self.value=0
 
def rowNum(p,sudoku):
	row=set(sudoku[p.y*9:(p.y+1)*9])
	row.remove(0)
	return row #set type
 
def colNum(p,sudoku):
	col=[]
	length=len(sudoku)
	for i in range(p.x,length,9):
		col.append(sudoku[i])
	col=set(col)
	col.remove(0)
	return col #set type
 
def blockNum(p,sudoku):
	block_x=p.x//3
	block_y=p.y//3
	block=[]
	start=block_y*3*9+block_x*3
	for i in range(start,start+3):
		block.append(sudoku[i])
	for i in range(start+9,start+9+3):
		block.append(sudoku[i])
	for i in range(start+9+9,start+9+9+3):
		block.append(sudoku[i])
	block=set(block)
	block.remove(0)
	return block #set type
 
def initPoint(sudoku):
	pointList=[]
	length=len(sudoku)
	for i in range(length):
		if sudoku[i]==0:
			p=point(i%9,i//9)
			for j in range(1,10):
				if j not in rowNum(p,sudoku) and j not in colNum(p,sudoku) and j not in blockNum(p,sudoku):
					p.available.append(j)
			pointList.append(p)
	return pointList
 
 
def tryInsert(p,sudoku):
	availNum=p.available
	for v in availNum:
		p.value=v
		if check(p,sudoku):
			sudoku[p.y*9+p.x]=p.value
			if len(pointList)<=0:
				t1=time.time()
				useTime=t1-t0
				showSudoku(sudoku)
				print('\nuse Time: %f s'%(useTime))
				exit()
			p2=pointList.pop()
			tryInsert(p2,sudoku)
			sudoku[p2.y*9+p2.x]=0
			sudoku[p.y*9+p.x]=0
			p2.value=0
			pointList.append(p2)
		else:
			pass	
 
def check(p,sudoku):
	if p.value==0:
		print('not assign value to point p!!')
		return False
	if p.value not in rowNum(p,sudoku) and p.value not in colNum(p,sudoku) and p.value not in blockNum(p,sudoku):
		return True
	else:
		return False
 
def showSudoku(sudoku):
	for j in range(9):
		for i in range(9):
			print('%d '%(sudoku[j*9+i]),end='')
		print('')	

调用求解主函数：

if __name__=='__main__':
	sudoku=[
			4,5,0,0,0,0,0,0,0,
			0,0,8,0,0,0,0,0,0,
			0,0,0,0,7,0,0,6,5,
			0,1,0,0,0,4,6,0,0,
			0,0,0,8,0,0,5,0,9,
			0,9,0,0,2,0,0,0,3,
			0,0,1,0,0,3,0,7,0,
			6,0,0,5,0,1,9,4,0,
			9,0,0,0,0,2,0,0,1,
			]
	pointList=initPoint(sudoku)
	showSudoku(sudoku)
	print('\n')
	p=pointList.pop()
	tryInsert(p,sudoku)

数独为空值的位置就是0，有数字的位置就写上，最后运行之后就能求解出来所有的位置的值了