以前写的一个八皇后问题求解，思路就是每次循环列出所有的可能解，然后过滤出不符合要求的解。详细见代码：

//检查两个点是否在攻击线上

def attack(p1,p2):

return p1[0]==p2[0] or p1[1] == p2[1] or (abs(p1[0]-p2[0])==abs(p1[1]-p2[1]))

//棋盘类

class Cell():

def __init__(self,n):

self.n = n

self.cell = [[(i,j) for j in range(n)] for i in range(n)]

def isAttack(p,rp):

for i in rp:

if attack(p,i):

return True

return False

def eightQueen(num):

cell = Cell(num)

board = cell.cell

result = [[(0,i)] for i in range(num)]

count = 0

for col in board:

count+=1

for p in col:

if not result:

result.append([p])

else:

#scan the result

for rp in result:

if not isAttack(p,rp):

copy = rp[:]

copy.append(p)

result.append(copy)

result = filter(lambda x: x.__len__()==count,result)

return filter(lambda x : x.__len__()==num,result)

//使用方式如下，返回所有的有效解

result = eightQueen(8)

下面是同事的递归方式的解决方案

class EightQueen:

#define const queen number

queenNum = 8

def __init__(self):

#定义一个8行 8 列的二维数组来模拟棋盘

self.qiPan = []

self.num = 0

#init a n*n list[][] as qi pan

for i in range(self.queenNum):

rows = []

for j in range(self.queenNum):

xy = [i,j]

rows.append(xy)

self.qiPan.append(rows)

#定义一个递归方法,递归去下一行寻找符合条件的坐标点

#param e8 存放符合条件的坐标点的列表

#param rowNum 寻找的行号

def getNext(self,e8,rowNum):

if len(e8) == self.queenNum:

self.num += 1

#输出符合条件的全部坐标点

print "method "+str(self.num)+"is:"+str(e8)

return

if rowNum > len(self.qiPan):

return

for d in self.qiPan[rowNum]:

if self.matchAllE8(e8,d):

e8copy = e8[:]

e8copy.append(d)

self.getNext(e8copy,rowNum+1)

def calcNum(self):

#遍历棋盘的第一行

for i in self.qiPan[0]:

#从第二行开始递归寻找符合条件的坐标

self.getNext([i],1)

print "all method is:"+str(self.num)

#判断left0坐标 是否满足上面行中已找到的全部符合条件的坐标都不攻击

def matchAllE8(self,e8,left0):

for i in e8:

if not self.isMatch(i,left0):

return False

return True

# 符合不攻击的条件

def isMatch(self,f,g):

return (f[0]+f[1] != g[0]+g[1]) and\

(f[0]-f[1] != g[0]-g[1]) and\

(f[0] != g[0]) and (f[1] != g[1])

e = EightQueen()

e.calcNum()