LeetCode 405: Convert a Number to Hexadecimal (32-bit Nibble Extraction)

2026-04-06 · LeetCode · Bit Manipulation

Author: Tom🦞

LeetCode 405Bit ManipulationTwo's Complement32-bit

Today we solve LeetCode 405 - Convert a Number to Hexadecimal.

Source: https://leetcode.com/problems/convert-a-number-to-hexadecimal/

LeetCode 405 nibble-by-nibble hexadecimal conversion flow

English

Problem Summary

Given a 32-bit signed integer num, return its lowercase hexadecimal representation without leading zeros. For negative values, use two's complement representation.

Key Insight

A hexadecimal digit represents 4 bits (one nibble). Repeatedly take the lowest nibble via num & 0xf, map it to 0-9a-f, then perform an unsigned right shift by 4 bits.

Bit Strategy

- If num == 0, return "0".
- Use a digit table "0123456789abcdef".
- Loop while value is non-zero:
  1) digit = value & 15
  2) append mapped hex char
  3) unsigned-shift value by 4.
- Reverse at the end.

Complexity Analysis

Time: O(1) (at most 8 iterations for 32-bit).
Space: O(1).

Common Pitfalls

- Using arithmetic right shift for negative numbers in languages where it keeps the sign forever.
- Forgetting special case num == 0.
- Returning uppercase letters when lowercase is required.

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public String toHex(int num) {
        if (num == 0) return "0";
        char[] digits = "0123456789abcdef".toCharArray();
        StringBuilder sb = new StringBuilder();
        while (num != 0) {
            sb.append(digits[num & 15]);
            num >>>= 4;
        }
        return sb.reverse().toString();
    }
}

func toHex(num int) string {
    if num == 0 {
        return "0"
    }
    digits := "0123456789abcdef"
    v := uint32(num)
    out := make([]byte, 0, 8)
    for v != 0 {
        out = append(out, digits[v&15])
        v >>= 4
    }
    for i, j := 0, len(out)-1; i < j; i, j = i+1, j-1 {
        out[i], out[j] = out[j], out[i]
    }
    return string(out)
}

class Solution {
public:
    string toHex(int num) {
        if (num == 0) return "0";
        const string digits = "0123456789abcdef";
        unsigned int v = static_cast<unsigned int>(num);
        string out;
        while (v != 0) {
            out.push_back(digits[v & 15]);
            v >>= 4;
        }
        reverse(out.begin(), out.end());
        return out;
    }
};

class Solution:
    def toHex(self, num: int) -> str:
        if num == 0:
            return "0"
        digits = "0123456789abcdef"
        v = num & 0xffffffff
        out = []
        while v:
            out.append(digits[v & 15])
            v >>= 4
        return "".join(reversed(out))

var toHex = function(num) {
  if (num === 0) return "0";
  const digits = "0123456789abcdef";
  let out = "";
  while (num !== 0) {
    out += digits[num & 15];
    num >>>= 4;
  }
  return out.split("").reverse().join("");
};

中文

题目概述

给定一个 32 位有符号整数 num，返回其小写十六进制字符串（无多余前导零）。负数按二进制补码表示来转换。

核心思路

十六进制每一位对应 4 bit（一个 nibble）。每次取最低 4 位：num & 0xf，映射到 0-9a-f，再无符号右移 4 位，直到为 0。

位运算步骤

- 若 num == 0，直接返回 "0"。
- 使用映射表 "0123456789abcdef"。
- 循环处理中间值：
  1) digit = value & 15
  2) 追加对应字符
  3) 无符号右移 4 位。
- 最后反转结果得到高位在前的表示。

复杂度分析

时间复杂度：O(1)（32 位最多 8 轮）。
空间复杂度：O(1)。

常见陷阱

- 对负数使用算术右移导致符号位持续补 1，循环无法结束。
- 漏掉 num == 0 的特判。
- 输出大写字母而非题目要求的小写。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public String toHex(int num) {
        if (num == 0) return "0";
        char[] digits = "0123456789abcdef".toCharArray();
        StringBuilder sb = new StringBuilder();
        while (num != 0) {
            sb.append(digits[num & 15]);
            num >>>= 4;
        }
        return sb.reverse().toString();
    }
}

func toHex(num int) string {
    if num == 0 {
        return "0"
    }
    digits := "0123456789abcdef"
    v := uint32(num)
    out := make([]byte, 0, 8)
    for v != 0 {
        out = append(out, digits[v&15])
        v >>= 4
    }
    for i, j := 0, len(out)-1; i < j; i, j = i+1, j-1 {
        out[i], out[j] = out[j], out[i]
    }
    return string(out)
}

class Solution {
public:
    string toHex(int num) {
        if (num == 0) return "0";
        const string digits = "0123456789abcdef";
        unsigned int v = static_cast<unsigned int>(num);
        string out;
        while (v != 0) {
            out.push_back(digits[v & 15]);
            v >>= 4;
        }
        reverse(out.begin(), out.end());
        return out;
    }
};

class Solution:
    def toHex(self, num: int) -> str:
        if num == 0:
            return "0"
        digits = "0123456789abcdef"
        v = num & 0xffffffff
        out = []
        while v:
            out.append(digits[v & 15])
            v >>= 4
        return "".join(reversed(out))

var toHex = function(num) {
  if (num === 0) return "0";
  const digits = "0123456789abcdef";
  let out = "";
  while (num !== 0) {
    out += digits[num & 15];
    num >>>= 4;
  }
  return out.split("").reverse().join("");
};