Question

我已经看了一会儿，我看不出我的二分搜索出了什么问题。如果我运行它，它会显示'RecursionError：比较超过最大递归深度'。任何人都可以看看下面有什么问题吗？谢谢！

#Write a function called in_bisect 
#that takes a sorted list and a target value 
#and returns the index
#of the value in the list if it’s there

def in_bisect(t, word):

    if len(t) == 0:
        return False

    middle = len(t) // 2

    if t[middle] == word:
        return middle

    elif t[middle] > word:
        return in_bisect(t[:middle], word)
    else:
        return in_bisect(t[middle:], word)


if __name__ == "__main__":
    fruits = ['apple', 'banana', 'kiwi', 'peach', 'watermelon']

    in_bisect(fruits, 'banana')
    in_bisect(fruits, 'ewf')

Answer 1

想象一下当列表长度为1时会发生什么。然后中间将为0.如果现在最终的else情况是实际情况，则递归将再次获得相同的列表（大小），具有相同的结果......无限递归。

解决方案：在middle添加一个，因为您已经知道middle本身不再是候选人：

else:
    return in_bisect(t[middle+1:], word)

Answer 2

我会在这里使用循环而不是递归，因为Python不能将尾递归转换为循环，并且对递归深度的限制非常低。

我还会检查False在def in_bisect(t, word): def iterator(start, end): # loop will terminate when exactly start = end - 1 while start < end - 1: middle = (start + end) // 2 if t[middle] == word: return middle elif t[middle] > word: end = middle else: start = middle + 1 # here we need to check wheither the last element in the list is the one we search for return start if t[start] == word else False # if len(t) is zero, our inner function would raise IndexError so we check it explicitly if len(t) == 0: return False return iterator(0, len(t))中的位置，否则返回.../legacy_pkg1/python/module1.py .../legacy_pkg1/python/module2.py .../legacy_pkg1/python/module3.py .../legacy_pkg2/python/module4.py .../legacy_pkg2/python/module5.py .../legacy_pkg2/python/module6.py。

PYTHONPATH=.../legacy_pkg1/python:.../legacy_pkg2/python

Answer 3

fruits = ['apple', 'banana', 'kiwi', 'peach', 'watermelon','dog','cat']

def in_bisect(t, word):
    # cheks if the word is in the list 
    if word not in t:
        return False 
    if len(t)==0:
        return False
    low=0
    high=len(fruits)
    #loop will when the value is found 
    while True:
        mid=(low + high)//2

        if t[mid]==word:

            return mid,t[mid]
        if word in t[:mid]  :
            high=mid
        else:
            low=mid
def test():
    print(in_bisect(fruits, 'apple'))
    print(in_bisect(fruits, 'banana'))
    print(in_bisect(fruits, 'kiwi'))
    print(in_bisect(fruits, 'dog'))
    print(in_bisect(fruits, 'ewf')   )
    print(in_bisect(fruits, 'banana'))
    return 'Test completed sucssfuly'
print (test())

python二分搜索练习

3 个答案: