二分法的两种实现（循环、递归及查找次数）

li=[i for i in range(10000000)]
s=23000
def compare(start=0,end=len(li),now=len(li)//2):
    if li[start]<s<li[now]:
        end = now
        now=(now+start)//2
        start=start
    elif li[now]<s<li[end]:
        start=now
        now = (now+end) // 2
        end=end
    elif s==li[start]:
        return start
    elif s==li[now]:
        return now
    elif s==li[end]:
        return end
    return compare(start,end,now)
print(compare())

li=[i for i in range(0,10)]
s=1
def compare(start=0,end=len(li)-1,now=(len(li)-1+0)//2,n=0):
    if li[start]<s<li[now]:
        end = now
        n+=1
        print(f'{n}次,{start}{now}{end}')
        now=(now+start)//2
        start=start
        n += 1
        print(f'{n}次,{start}{now}{end}')
    elif li[now]<s<li[end]:
        start=now
        n += 1
        print(f'{n}次,{start}{now}{end}')
        now = (now+end) // 2
        end=end
        n += 1
        print(f'{n}次,{start}{now}{end}')
    elif s==li[start]:
        n += 1
        print(f'{n}次,{start}{now}{end}')
        return start,n
    elif s==li[now]:
        n += 1
        print(f'{n}次,{start}{now}{end}')
        return now,n
    elif s==li[end]:
        n += 1
        print(f'{n}次,{start}{now}{end}')
        return end,n
    return compare(start,end,now,n)
print(compare())

li=[i for i in range(100)]
s=23
start=0
end=100
now=50
while 1 :
    if li[start]<s<li[now]:
        end = now
        now=(now+start)//2
        start=start
    elif li[now]<s<li[end]:
        start=now
        now = (now+end) // 2
        end=end
    elif s==li[start]:
        m= start
        break
    elif s==li[now]:
        m=now
        break
    elif s==li[end]:
        m= end
        break
print(m)