python奇怪行为中的二进制搜索

本文介绍了python奇怪行为中的二进制搜索的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

请查看以下代码:

def chop(array, search): lo = 0 high = len(array) - 1 while lo <= high: mid = (high + lo) /2 if array[mid] == search: return 'true' elif search > array[mid]: low = mid + 1 else: high = mid - 1 return 'false' if __name__ == '__main__': a = [1,2,3,4,5,6,7,8,9,10] print chop(a, 3)

我写了这个小脚本，该脚本应该在数组中搜索数字-常规二进制搜索.因此，我运行了脚本，例如，当我放入chop(a, 1)时，我就得到了真，而当我放入chop(a, 2)时，我就得到了，但是当我放入chop(a, 3)时，我没有得到任何答案，只是在其中插入了空行Python Shell.

I wrote this little script which is supposed to search for number in array - regular binary search. So I run the script, and for example when I put in chop(a, 1) I get true, when I put in chop(a, 2) I get true, but when I put chop(a, 3) I don't get an answer, just empty line in the Python Shell.

有人对发生的事情有想法吗?

Does anyone have an idea on what is going on?

推荐答案

我猜这是您的错误:

low = mid + 1

您的while循环使用变量lo，并且您正在while循环中定义一个名为low的新变量.本质上，您永远不会更新lo变量.

Your while loop uses the variable lo, and you're defining a new variable called low within your while loop. In essence, you're never updating your lo variable.

将该行更改为:

lo = mid + 1

您的算法应该可以使用.

and your algorithm should work.