八皇后问题（Python实现）

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看《Python基础教程》，看到生成器一章，提到八皇后问题，没有继续往后看，自己试着写了一个。

八皇后问题是数据结构里面的经典问题，思路主要是利用回溯法，利用栈保留走过的路径，走过的路，入栈，走不通了，出栈，继续往下尝试。

思路比较简单，写个伪码没什么问题。但是写得时间比较长，要是面试的时候让你笔写个无BUG可运行版本，那还真得喊救命了

运行结果：能够打印出全部92种解法，只是判断冲突的办法，太low了...................

#encoding:utf-8import copydef checkAttack(array):for i in range(len(array)):x=sum(array[i])check="水平"+str(x)if (x>1):print checkreturn Truefor i in range(len(array)):x=[]for j in range(len(array)):x.append(array[j][i])y=sum(x)check="竖直"+str(y)if(y>1):print checkreturn Truefor  i in range(len(array)):x=[]p=[]j=ik=0while j>=0 and k<=i:x.append(array[j][k])p.append(array[len(array)-1-k][len(array)-1-j])j-=1k+=1y=sum(x)q=sum(p)check="左对"+str((y,q))if y>1 or q>1:print checkreturn Truefor  i in range(len(array)-1,-1,-1):x=[]p=[]j=ik=0while j<=len(array)-1:x.append(array[k][j])p.append(array[j][k])j+=1k+=1y=sum(x)q=sum(p)check="右对"+str((y,q))if y>1 or q>1:print checkreturn Truereturn Falsedef initBorad():board=[];for _ in range(8):x=map( lambda x:x-x,range(8) )board.append(x)return boarddef eightQueen():board=initBorad()allBoard=[]stack=[]i=j=0try:while True:while  i < len(board):print (i,j)found=Falsewhile j < len(board):board[i][j]=1if (checkAttack(board)==True):board[i][j]=0j+=1else:stack.append((i,j))print "in stack"+str((i,j))found=Truei+=1j=0breakif found==False:i,j=stack.pop()board[i][j]=0print "out stack"+str((i,j)) j+=1for xp in board:print xpallBoard.append(board)print "目前找到："+str(len(allBoard))x=copy.deepcopy(board)board=xi,j=stack.pop()board[i][j]=0j+=1except (Exception) as e:print ereturn allBoardx=eightQueen()count=0for ie in x:count+=1print "\n解法:"+str(count)for ii in ie:print ii

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