如何解决Python列表中的二进制搜索
好吧,您的代码中有些小错误。要找到它们,您应该使用调试器,或者至少添加跟踪以了解会发生什么。这是您的原始代码,带有使问题显而易见的跟踪:
def search(list, target):
min = 0
max = len(list)-1
avg = (min+max)/2
print list, target, avg
...
您可以立即看到:
修复程序现在很简单:
elif (list[avg] < target):
return avg + 1 + search(list[avg+1:], target) # add the offset
else:
return search(list[:avg], target) # sublist ends below the upper limit
这还不是全部,当您以结束循环时min == max
,您不会返回任何内容(意味着您将返回None)。最后但并非最不重要的一点是,
为自己的变量使用标准Python库中的名称。
所以这是固定代码:
def search(lst, target):
min = 0
max = len(lst)-1
avg = (min+max)/2
# uncomment next line for traces
# print lst, target, avg
while (min < max):
if (lst[avg] == target):
return avg
elif (lst[avg] < target):
return avg + 1 + search(lst[avg+1:], target)
else:
return search(lst[:avg], target)
# avg may be a partial offset so no need to print it here
# print "The location of the number in the array is", avg
return avg
解决方法
我正在尝试在python中的列表上执行二进制搜索。列表是使用命令行参数创建的。用户输入要在数组中查找的数字,然后返回该元素的索引。由于某种原因,该程序仅输出1和无。代码如下。任何帮助都非常感谢。
import sys
def search(list,target):
min = 0
max = len(list)-1
avg = (min+max)/2
while (min < max):
if (list[avg] == target):
return avg
elif (list[avg] < target):
return search(list[avg+1:],target)
else:
return search(list[:avg-1],target)
print "The location of the number in the array is",avg
# The command line argument will create a list of strings
# This list cannot be used for numeric comparisions
# This list has to be converted into a list of ints
def main():
number = input("Please enter a number you want to search in the array !")
index = int(number)
list = []
for x in sys.argv[1:]:
list.append(int(x))
print "The list to search from",list
print(search(list,index))
if __name__ == '__main__':
main()
CL :
Anuvrats-MacBook-Air:Python anuvrattiku$ python binary_search.py 1 3 4 6 8 9 12 14 16 17 27 33 45 51 53 63 69 70
Please enter a number you want to search in the array !69
The list to search from [1,3,4,6,8,9,12,14,16,17,27,33,45,51,53,63,69,70]
0
Anuvrats-MacBook-Air:Python anuvrattiku$
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。