Python中的二进制搜索实现

本文介绍了Python中的二进制搜索实现的处理方法，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

我正在尝试使用二进制搜索实现解决方案.我有一个数字列表

I am trying to implement a solution using binary search. I have a list of numbers

list = [1, 2, 3, 4, 6] value to be searched = 2

我写了这样的东西

def searchBinary(list, sval): low = 0 high = len(list) while low < high: mid = low + math.floor((high - low) / 2) if list[mid] == sval: print("found : ", sval) elif l2s[mid] > sval: high = mid - 1 else: low = mid + 1

但是当我尝试实现这一点时，出现了类似错误:索引超出范围.请帮助确定问题.

but when I am trying to implement this, I am getting an error like: index out of range. Please help in identifying the issue.

推荐答案

一些事情.

您的命名不一致.另外，请勿将list用作变量名，否则会掩盖全局内置变量.

Your naming is inconsistent. Also, do not use list as a variable name, you're shadowing the global builtin.

停止条件为while low <= high.这个很重要.

The stopping condition is while low <= high. This is important.

找到值时不会中断.这将导致无限递归.

You do not break when you find a value. This will result in infinite recursion.

def searchBinary(l2s, sval): # do not use 'list' as a variable low = 0 high = len(l2s) while low <= high: # this is the main issue. As long as low is not greater than high, the while loop must run mid = (high + low) // 2 if l2s[mid] == sval: print("found : ", sval) return elif l2s[mid] > sval: high = mid - 1 else: low = mid + 1

现在，

list_ = [1, 2, 3, 4, 6] searchBinary(list_, 2)

输出:

found : 2