八皇后问题 （循环递归法）

#* queen problem with recurison
BOARD_SIZE = 8

def under_attack(col, queens):
   left = right = col
   for r, c in reversed(queens):
 #左右有冲突的位置的列号
       left, right = left - 1, right + 1

       if c in (left, col, right):
           return True
   return False

def solve(n):
   if n == 0:
       return [[]]

   smaller_solutions = solve(n - 1)

   return [solution+[(n,i+1)]
       for i in xrange(BOARD_SIZE)
           for solution in smaller_solutions
               if not under_attack(i+1, solution)]
for answer in solve(BOARD_SIZE):
   print answer

改进排序数组的索引：

def deduplication(self, nums):
    if self in nums:                  #加入数字是否在数组的判断，不在则打印none
        for i in range(len(nums)):    #一个for循环加上if else即可找出数组索引
            if nums[i] == self:
                return i
            else:
                i += 1
        return i

print(deduplication(5, [1, 3, 5, 6])