我想进行二进制搜索,但结果有误

I wanted to do a Binary search but the result is faulty

我想对列表进行二分查找,但结果显示 'false',即使我检查了列表中的数字。

def clist(a):

    l = [2,6,5,9,7,1,4,8,3]
    newl = sorted(l)
    check = int(1+len(newl)/2)

    if newl[check] == a:
        return True

    if check > a:
        for x in newl[:check]:
            if x == a:
                return True
            return False

    if check < a:
        for x in newl[check::]:
            if x == a:
                return True
            return False

print(clist(7))

请仔细阅读:

def clist(a):

    l = [2,6,5,9,7,1,4,8,3]
    newl = sorted(l)
    check = int(1+len(newl)/2)

    if newl[check] == a:
        return True

    if newl[check] > a:              #fixed the bug here
        for x in newl[:check]:
            if x == a:
                return True

    if newl[check] < a:             #fixed the bug here
        for x in newl[check:]:
            if x == a:
                return True

    return False        #Return false should be universal. When the entire search fails it should be called.

print(clist(7))

你的函数不是二分查找,你是在检查了中间元素之后逐个元素地检查排序列表。

def binary_search(arr, i):
    n = len(arr)
    arr = sorted(arr)
    left = 0
    right = n - 1

    # Define the condition when the loop should be broken 
    while (left <= right):
        mid = left + (right-left) // 2
        if arr[mid] == i:
            return True
        elif arr[mid] < i:
            left = mid + 1 
        else:
            right = mid - 1
    return False

l = [2,6,5,9,7,1,4,8,3]
i = 7
binary_search(l, i)

您可以这样编写脚本:

  1. 取列表中间的元素
  2. return 如果这就是你需要的
  3. 如果你的needle比中间的gt,那么在列表的剩余右侧调用bsearch
  4. 另外用左边叫bsearch
def bsearch(needle, haystack):
    l = len(haystack)
    half = int(l / 2)
    element = haystack[half];

    if element == needle:
        return element

    if needle <= element:
        return bsearch(needle, haystack[0:half])

    if needle > element:
        return bsearch(needle, haystack[half:l])




print(bsearch(7, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]))

在二进制搜索中:

  1. 列表必须排序
  2. 如@tripleee所述,您必须递归地将列表分成两半