时间复杂度:O(logn)
# _*_coding:utf-8_*_
def binary_search(li, val):
left = 0
right = len(li) - 1
while left <= right: # 候选区有值
mid = (left + right) // 2
if li[mid] == val:
return mid
elif li[mid] > val: # 带查找的值在mid左侧
right = mid - 1
else: # li[mid] < val 带查找的值在mid右侧
left = mid + 1
else:
return None
li = list(range(1000))
print(binary_search(li, 389))