405. Convert a Number to Hexadecimal

原创于 2017-02-01 08:20:43 发布 · 343 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode #Bit manipulation #Convert a Number to

LeetCode 同时被 2 个专栏收录

402 篇文章

订阅专栏

Bit manipulation

24 篇文章

订阅专栏

本文介绍了一种将整数转换为十六进制字符串的算法，特别针对负数使用了二进制补码的方法。该算法不依赖于任何直接进行格式化的库函数，并提供了一个Java实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Note:

All letters in hexadecimal (a-f) must be in lowercase.
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.
The given number is guaranteed to fit within the range of a 32-bit signed integer.
You must not use any method provided by the library which converts/formats the number to hex directly.

Example 1:

Input:
26

Output:
"1a"

Example 2:

Input:
-1

Output:
"ffffffff"

10进制转化2进制，每四位一个组合，如果是0，返回‘0’。代码如下：

public class Solution {
    char[] ch = {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};
    
    public String toHex(int num) {
        if (num == 0) {
            return "0";
        }
        StringBuilder result = new StringBuilder();
        while (num != 0) {
            result.insert(0, ch[num & 15]);
            num = num >>> 4;
        }
        return result.toString();
    }
}