leetcode 第十六周解题报告

最新推荐文章于 2025-08-07 15:32:36 发布

原创最新推荐文章于 2025-08-07 15:32:36 发布 · 230 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法

本文介绍了一种不使用库函数直接转换整数到十六进制的方法。通过位操作提取整数的每一位，并将其组织成十六进制形式。特别地，对于负数采用二进制补码的方式进行处理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

问题描述：

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.

解题思路：

用移位操作将数的每一位取出保存在bool数组里面，再按每四位为一个单元的顺序重新组织十六进制数即可。

代码：

#include<cmath>
class Solution {
public:
    string toHex(int num) {
        char codeT[] = {'0', '1', '2', '3', '4','5',
            '6','7','8','9','a','b','c','d','e','f'
        };
        if (num == 0) return string("0");
        //bool sign = (num > 0 ? true : false);
        bool bits[32];
        for(int i = 31; i >= 0; --i){
            bits[31 - i] = num & (1 << i);
        }
        string res("00000000");
        for(int i = 0; i < 8; ++i){
            int tmp = 0;
            for(int j = 0; j < 4; ++j){
                tmp = (tmp << 1) + (int)(bits[i * 4 + j]);
            }
            res[i] = codeT[tmp];
        }
        for(int i = 0; i < 8; ++i){
            if(res[i] != '0'){
                return std::string(res.begin() + i, res.end());
            }
        }
    }
};