我想对列表进行二进制搜索,但是即使我检查列表中的数字,结果也显示为“ false”。
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))
答案 0 :(得分:2)
您可以使用以下方式编写脚本:
needle
位于中间,则请在列表的其余右侧调用bsearch
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:
答案 1 :(得分:1)
请经历以下步骤:
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))
答案 2 :(得分:0)
您的函数不是二进制搜索,您要在检查中间元素之后逐个检查排序列表中的元素。
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)