如何在没有库函数的情况下将字符串解析为整数？

本文介绍了如何在没有库函数的情况下将字符串解析为整数？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

我最近在一次采访中被问到这个问题：

I was recently asked this question in an interview:

如何在不使用任何库的情况下将12345形式的字符串解析为其整数表示12345功能，而不管语言？

"How could you parse a string of the form '12345' into its integer representation 12345 without using any library functions, and regardless of language?"

我想到了两个答案，但面试官说有三分之一。以下是我的两个解决方案：

I thought of two answers, but the interviewer said there was a third. Here are my two solutions:

解决方案1：保留一个字典，其中包含'1'=> 1，'2'=> 2等。然后解析字符串一一次一个字符，查找字典中的字符，然后乘以位置值。求和结果。

Solution 1: Keep a dictionary which maps '1' => 1, '2' => 2, etc. Then parse the string one character at a time, look up the character in your dictionary, and multiply by place value. Sum the results.

解决方案2：一次解析一个字符，并从每个字符中减去0。这将给你'1' - '0'= 0x1，'2' - '0'= 0x2等。再次，乘以位值并对结果求和。

Solution 2: Parse the string one character at a time and subtract '0' from each character. This will give you '1' - '0' = 0x1, '2' - '0' = 0x2, etc. Again, multiply by place value and sum the results.

有人能想到第三种解决方案可能是什么吗？

Can anyone think of what a third solution might be?

谢谢。

推荐答案

我希望这是面试官所追求的：

I expect this is what the interviewer was after:

number = "12345" value = 0 for digit in number: //Read most significant digit first value = value * 10 + valueOf(digit)

此方法使用的操作远少于您概述的方法。

This method uses far less operations than the method you outlined.