Convert a Number to Hexadecimal

2023-12-27 04:58
文章标签 number convert hexadecimal

本文主要是介绍Convert a Number to Hexadecimal，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目地址：https://leetcode.com/problems/convert-a-number-to-hexadecimal/

Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, two’s complement method is used.

Note:

  1. All letters in hexadecimal (a-f) must be in lowercase.
  2. The hexadecimal string must not contain extra leading 0s. If the number is zero, it is represented by a single zero character ‘0’; otherwise, the first character in the hexadecimal string will not be the zero character.
  3. The given number is guaranteed to fit within the range of a 32-bit signed integer.
  4. You must not use any method provided by the library which converts/formats the number to hex directly.

Example 1: 

Input:
26Output:
"1a"

Example 2:

Input:
-1Output:
"ffffffff"

题目是将十进制转换为十六进制，整数还是比较好处理的，但是这个题目的关键是负数，要理解负数在计算机中是如何表示的：

public class ConvertANumberToHexadecimal {public static final String[] digit= {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f"};public String toHex(int num) {StringBuilder reslut = new StringBuilder("");if (num > 0) {while (num != 0) {reslut.insert(0, digit[num % 16]);num /= 16;}} else if (num == 0){reslut.insert(0, "0");} else {// 以下是截取十六进制每位对应的二进制数值// 15 to binnary: 0000 0000 0000 0000 0000 0000 0000 1111reslut.insert(0, digit[num & 15]);// 240 to binnary: 0000 0000 0000 0000 0000 0000 1111 0000reslut.insert(0, digit[(num & 240) >> 4]);// 3840 to binnary: 0000 0000 0000 0000 0000 1111 0000 0000reslut.insert(0, digit[(num & 3840) >> 8]);// 61440 to binnary: 0000 0000 0000 0000 1111 0000 0000 0000reslut.insert(0, digit[(num & 61440) >> 12]);// 983040 to binnary: 0000 0000 0000 1111 0000 0000 0000 0000reslut.insert(0, digit[(num & 983040) >> 16]);// 15728640 to binnary: 0000 0000 1111 0000 0000 0000 0000 0000reslut.insert(0, digit[(num & 15728640) >> 20]);// 251658240 to binnary: 0000 1111 0000 0000 0000 0000 0000 0000reslut.insert(0, digit[(num & 251658240) >> 24]);// 1879048192 to binnary: 0111 0000 0000 0000 0000 0000 0000 0000reslut.insert(0, digit[((num & 1879048192) >> 28) + 8]);}return reslut.toString();}public static void main(String[] args) {ConvertANumberToHexadecimal hexadecimal = new ConvertANumberToHexadecimal();System.out.println(hexadecimal.toHex(-13));}
}

其实在JDK中已经封装好了十进制转十六进制的方法，它是这么写的：

Integer.toHexString(int i)

那么我们看一下JDK是怎么实现的：

/*** Returns a string representation of the integer argument as an* unsigned integer in base&nbsp;16.** <p>The unsigned integer value is the argument plus 2<sup>32</sup>* if the argument is negative; otherwise, it is equal to the* argument.  This value is converted to a string of ASCII digits* in hexadecimal (base&nbsp;16) with no extra leading* {@code 0}s.** <p>The value of the argument can be recovered from the returned* string {@code s} by calling {@link* Integer#parseUnsignedInt(String, int)* Integer.parseUnsignedInt(s, 16)}.** <p>If the unsigned magnitude is zero, it is represented by a* single zero character {@code '0'} ({@code '\u005Cu0030'});* otherwise, the first character of the representation of the* unsigned magnitude will not be the zero character. The* following characters are used as hexadecimal digits:** <blockquote>*  {@code 0123456789abcdef}* </blockquote>** These are the characters {@code '\u005Cu0030'} through* {@code '\u005Cu0039'} and {@code '\u005Cu0061'} through* {@code '\u005Cu0066'}. If uppercase letters are* desired, the {@link java.lang.String#toUpperCase()} method may* be called on the result:** <blockquote>*  {@code Integer.toHexString(n).toUpperCase()}* </blockquote>** @param   i   an integer to be converted to a string.* @return  the string representation of the unsigned integer value*          represented by the argument in hexadecimal (base&nbsp;16).* @see #parseUnsignedInt(String, int)* @see #toUnsignedString(int, int)* @since   JDK1.0.2*/
public static String toHexString(int i) {
    return toUnsignedString0(i, 4);
}

然后再看下toUnsignedString0及其内部所涉及到的方法的实现：

/*** Convert the integer to an unsigned number.*/
private static String toUnsignedString0(int val, int shift) {// assert shift > 0 && shift <=5 : "Illegal shift value";int mag = Integer.SIZE - Integer.numberOfLeadingZeros(val);int chars = Math.max(((mag + (shift - 1)) / shift), 1);char[] buf = new char[chars];formatUnsignedInt(val, shift, buf, 0, chars);// Use special constructor which takes over "buf".return new String(buf, true);
}/*** Returns the number of zero bits preceding the highest-order* ("leftmost") one-bit in the two's complement binary representation* of the specified {@code int} value.  Returns 32 if the* specified value has no one-bits in its two's complement representation,* in other words if it is equal to zero.** <p>Note that this method is closely related to the logarithm base 2.* For all positive {@code int} values x:* <ul>* <li>floor(log<sub>2</sub>(x)) = {@code 31 - numberOfLeadingZeros(x)}* <li>ceil(log<sub>2</sub>(x)) = {@code 32 - numberOfLeadingZeros(x - 1)}* </ul>** @param i the value whose number of leading zeros is to be computed* @return the number of zero bits preceding the highest-order*     ("leftmost") one-bit in the two's complement binary representation*     of the specified {@code int} value, or 32 if the value*     is equal to zero.* @since 1.5*/
public static int numberOfLeadingZeros(int i) {// HD, Figure 5-6if (i == 0)return 32;int n = 1;if (i >>> 16 == 0) { n += 16; i <<= 16; }if (i >>> 24 == 0) { n +=  8; i <<=  8; }if (i >>> 28 == 0) { n +=  4; i <<=  4; }if (i >>> 30 == 0) { n +=  2; i <<=  2; }n -= i >>> 31;return n;
}/*** Format a long (treated as unsigned) into a character buffer.* @param val the unsigned int to format* @param shift the log2 of the base to format in (4 for hex, 3 for octal, 1 for binary)* @param buf the character buffer to write to* @param offset the offset in the destination buffer to start at* @param len the number of characters to write* @return the lowest character  location used*/static int formatUnsignedInt(int val, int shift, char[] buf, int offset, int len) {int charPos = len;int radix = 1 << shift;int mask = radix - 1;do {buf[offset + --charPos] = Integer.digits[val & mask];val >>>= shift;} while (val != 0 && charPos > 0);return charPos;
}/*
* Package private constructor which shares value array for speed.
* this constructor is always expected to be called with share==true.
* a separate constructor is needed because we already have a public
* String(char[]) constructor that makes a copy of the given char[].
*/
String(char[] value, boolean share) {// assert share : "unshared not supported";this.value = value;
}

这篇关于Convert a Number to Hexadecimal的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！


http://www.chinasem.cn/article/541938。

相关文章

usaco 1.2 Name That Number（数字字母转化）

usaco 1.2 Name That Number（数字字母转化）

巧妙的利用code[b[0]-'A'] 将字符ABC...Z转换为数字 需要注意的是重新开一个数组 c [ ] 存储字符串 应人为的在末尾附上 ‘ \ 0 ’ 详见代码： /*ID: who jayLANG: C++TASK: namenum*/#include<stdio.h>#include<string.h>int main(){FILE *fin = fopen (
阅读更多...
题目1380：lucky number

题目1380：lucky number

题目1380：lucky number 时间限制：3 秒 内存限制：3 兆 特殊判题：否 提交：2839 解决：300 题目描述： 每个人有自己的lucky number，小A也一样。不过他的lucky number定义不一样。他认为一个序列中某些数出现的次数为n的话，都是他的lucky number。但是，现在这个序列很大，他无法快速找到所有lucky number。既然
阅读更多...
Jenkins 通过 Version Number Plugin 自动生成和管理构建的版本号

Jenkins 通过 Version Number Plugin 自动生成和管理构建的版本号

步骤 1：安装 Version Number Plugin 登录 Jenkins 的管理界面。进入 “Manage Jenkins” -> “Manage Plugins”。在 “Available” 选项卡中搜索 “Version Number Plugin”。选中并安装插件，完成后可能需要重启 Jenkins。 步骤 2：配置版本号生成 打开项目配置页面。在下方找到 “Build Env
阅读更多...
【Hdu】Minimum Inversion Number(逆序，线段树)

【Hdu】Minimum Inversion Number(逆序，线段树)

利用线段树在nlogn的时间复杂度内求一段数的逆序。 由于给的序列是由0 ~ n -1组成的，求出初始的逆序之后可以递推出移动之后的逆序数。 #include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const in
阅读更多...
【JavaScript】基本数据类型与引用数据类型区别（及为什么String、Boolean、Number基本数据类型会有属性和方法？）

【JavaScript】基本数据类型与引用数据类型区别（及为什么String、Boolean、Number基本数据类型会有属性和方法？）

基本数据类型 　　JavaScript基本数据类型包括：undefined、null、number、boolean、string。基本数据类型是按值访问的，就是说我们可以操作保存在变量中的实际的值。 1）基本数据类型的值是不可变的 任何方法都无法改变一个基本类型的值，比如一个字符串： var name = "change";name.substr();//hangconsole.log
阅读更多...
ORA-24067: exceeded maximum number of subscribers for queue ADMIN.SMS_MT_QUEUE

ORA-24067: exceeded maximum number of subscribers for queue ADMIN.SMS_MT_QUEUE

临时处理办法： delete from aq$_ss_MT_tab_D;delete from aq$_ss_MT_tab_g;delete from aq$_ss_MT_tab_h;delete from aq$_ss_MT_tab_i;delete from aq$_ss_MT_tab_p;delete from aq$_ss_MT_tab_s;delete from aq$
阅读更多...
org.springframework.core.convert.ConversionFailedException

org.springframework.core.convert.ConversionFailedException

适用： 1.你在springmvc.xml中写了静态资源处理<mvc:resources location="/" mapping="/**"></mvc:resources> 2.你在springmvc.xml中写了日期类型转换 解决方法: 首先检测你的spring版本，然后 1.spring3.0后将<mvc:resources location="/" mapping="/**">
阅读更多...
SQLSERVER排名函数RANK，DENSE_RANK，NTILE，ROW_NUMBER

SQLSERVER排名函数RANK，DENSE_RANK，NTILE，ROW_NUMBER

SQL SERVER排名函数RANK，DENSE_RANK，NTILE，ROW_NUMBER 前言 本文意于用实例数据帮助理解SQL SERVER排名函数RANK，DENSE_RANK，NTILE，ROW_NUMBER。 准备工作 创建测试表：   ? 1 2 3 4 5 create table test( id int identity(1,1)
阅读更多...
[LeetCode] 137. Single Number II

[LeetCode] 137. Single Number II

题：https://leetcode.com/problems/single-number-ii/ 题目大意 给定array，其中有一个元素只出现了1次，其他元素都出现了3次。 思路 求和 减去 (set(array)*3 - array)/2 作为答案。 class Solution {public int singleNumber(int[] nums) {Set<Long> se
阅读更多...
Oracle - ORA-01789: Query block has incorrect number of result columns

Oracle - ORA-01789: Query block has incorrect number of result columns

一、原因     这个错误一般是在执行表之间的相加（union），相减（minus）等SQL语句时，两个个查询块具有不一致的结果列数所导致的。 二、方案     只要将两段SQL语句的列数调整为一致就可以解决。使用union时，要注意数据库字段的格式要一致，如varchar和nvarchar是不一样的。
阅读更多...

Copyright China编程 23002807@qq.com

沪ICP备2023033072号-2