有的公司面试喜欢问这个问题，问题是这样的，12个小球有1个重量不一样，给一个天平，最多称3次，把这个重量不一样的小球找出来
解决这个问题比较难想到的是天平两边放4个球，重量一样的情况，需要怎么调整球的位置。
我把解决方法整理为python代码，如下：
# 左边重
LEFT = '<'
# 右边重
RIGHT = '>'
# 左右相等
EQUAL = '='
def weight(left_balls, right_balls):
"""
天平对两边的球称重
:param left_balls: 天平左边的球
:param right_balls: 天平右边的球
:return:
"""
if not isinstance(left_balls, list):
left_balls = [left_balls]
if not isinstance(right_balls, list):
right_balls = [right_balls]
if sum(left_balls) < sum(right_balls):
return LEFT
elif sum(left_balls) == sum(right_balls):
return EQUAL
else:
return RIGHT
def find_different_in_12(balls):
result = weight(balls[:4], balls[4: 8])
if result == EQUAL:
result = weight(balls[:3], balls[8: 11])
if result == EQUAL:
return 12, weight(balls[-1], balls[0])
elif result == LEFT:
result = weight(balls[8], balls[9])
if result == EQUAL:
return 11, '-'
elif result == LEFT:
return 10, '>'
else:
return 9, '>'
elif result == LEFT:
# 关键的地方
result = weight([balls[4], balls[1], balls[8], balls[9]], [balls[0], balls[5], balls[6], balls[10]])
if result == EQUAL:
result = weight([balls[2], balls[7]], [balls[8], balls[9]])
if result == EQUAL:
return 3, '<'
elif result == RIGHT:
return 7, '>'
elif result == LEFT:
return 2, '<'
elif result == LEFT:
result = weight([balls[1], balls[5]], [balls[8], balls[9]])
if result == EQUAL:
return 6, '>'
elif result == LEFT:
return 1, '<'
elif result == RIGHT:
return 5, '>'
elif result == RIGHT:
result = weight(balls[0], balls[8])
if result == EQUAL:
return 4, '>'
elif result == LEFT:
return 0, '<'
elif result == RIGHT:
# 省略，跟result == LEFT类似
pass
return 'no result'
def get_balls(i, w):
"""
把第i个球换成重量为w的不一样的球
:param i:
:param w:
:return:
"""
balls = [1] * 12
balls[i] = w
print(balls)
return balls
if __name__ == "__main__":
balls = get_balls(8, 2)
print(find_different_in_12(balls))
balls = get_balls(6, 2)
print(find_different_in_12(balls))
执行后可以看到结果：
[1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1]
(9, '>')
[1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1]
(6, '>')
