python算法——第四天

def func(num):
if num / 2 > 0:
num -= 1
print(num)
num = func(num)
print('quit')
return num

func(10)

def binary_search(data_list,find_num):
mid_pos = int(len(data_list) / 2)  # 计算需要查找数据的长度的一半
mid_val = data_list[mid_pos]  # 获取中间位置的那个值
print(data_list)  # 查看每次剩余筛选的数据列表
if len(data_list) > 0:  # 当列表长度大于0时,则一直查找
if mid_val > find_num:  # 如果中间的数比实际要查找的数大,那么这个数肯定在左边
print("%s should be in left of [%s]" % (find_num, mid_val))
binary_search(data_list[:mid_pos], find_num)
elif mid_val < find_num:  # 如果中间的数比实际查找的数小,那么这个数肯定在右边
print("%s should be in right of [%s]" % (find_num, mid_val))
binary_search(data_list[mid_pos:], find_num)
else:  # 如果中间数与实际查找的数恰巧相等,那么这个数肯定是要找的拿个数
print("Find %s" % find_num)

else:  # 否则就是买药这个数
print("cannot find [%s] in data_list" % find_num)

if __name__ == '__main__':
primes = [1, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
binary_search(primes, 1)   # 在列表里面查找1

array = [[col for col in range(4)] for row in range(4)]  # 初始化一个4*4数组
# array=[[col for col in 'abcd'] for row in range(4)]

for row in array:  # 旋转前先看看数组长啥样
print(row)

for i, row in enumerate(array):

for index in range(i, len(row)):
tmp = array[index][i]  # 将每一列数据在每一次遍历前,临时存储
array[index][i] = array[i][index]  # 将每一次遍历行的值,赋值给交叉的列
print(tmp, array[i][index])  # = tmp
array[i][index] = tmp  # 将之前保存的交叉列的值,赋值给交叉行的对应值
for r in array:  # 打印每次交换后的值
print(r)