day14 Python 二分法的三种实现

一、使用循环实现二分法

n = int(input("plz input your num: "))
lst = [4, 56, 178, 253, 625, 1475, 2580, 3574, 15963]

left = 0
right = len(lst) - 1

while left <= right:
    center = (left + right) // 2
    if n > lst[center]:
        left = center + 1
    elif n < lst[center]:
        right = center - 1
    else:
        print("I'am find it!")
        break
else:
    print("it's not exist!")

二、通过修改列表下标来实现二分法

lst = [4, 56, 178, 253, 625, 1475, 2580, 3574, 15963]


def findN(n, left=0, right=(len(lst) - 1), retry=0):
    center = (left + right) // 2
    print(f"search --> N:{n}, left:{left}, right:{right}, center:{center}, retry:{retry}")
    if left <= right:
        if n < lst[center]:
            right = center - 1
            retry += 1
            return findN(n, left, right, retry)
        elif n > lst[center]:
            retry += 1
            left = center + 1
            return findN(n, left, right, retry)
        else:
            print("I'am find it!")
            return True
    else:
        print("it's not exist!")

print(findN(253))

三、通过列表切片来实现二分法

lst = [4, 56, 178, 253, 625, 1475, 2580, 3574, 15963]

def findN(lst,n):
    center = len(lst) // 2

    if len(lst) > 0:
        if n > lst[center]:
            return findN(lst[center+1:],n)
        elif n < lst[center]:
            return findN(lst[:center],n)
        else:
            print(f"Num {n}, I'am find it!")
            return True
    else:
        print(f"Num {n}, It's not exist!")

findN(lst,666)