Question

我有一个程序，它使用二进制搜索。最后，我需要打印一个循环计数。如何做得更好？

import re

def binarySearch(sumList, whattofind):
    a=0
    if len(sumList) == 0:
        return False
    else:
        midpoint = len(sumList)/2
        if sumList[midpoint]==whattofind:
            a=a+1
            print(a)
            return True
        else:
            if whattofind<sumList[midpoint]:
                a+=1
                return binarySearch(sumList[:midpoint],whattofind)
            else:
                a+=1
                return binarySearch(sumList[midpoint+1:],whattofind)
        print(a)
result = re.findall(r'\w\w', open("text.txt","r").read())
sumList=[]
for line in result:
    sumList.append(ord(line[0])+ord(line[1]))
sumList.sort()
whattofind=int(input('Enter number: '))
print (sumList)
print(binarySearch(sumList, whattofind))

Answer 1

执行以下操作

count = 0
def binarySearch(sumList, whattofind):
    global count
    count += 1

并且在最后一行代码中只打印count

的值

如何计算周期并打印它

1 个答案: