LeetCode 405 - Convert a Number to Hexadecimal

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:
26

Output:
"1a"

Example 2:

Input:
-1

Output:
"ffffffff"

https://discuss.leetcode.com/topic/60365/simple-java-solution-with-comment

Basic idea: each time we take a look at the last four digits of
            binary verion of the input, and maps that to a hex char
            shift the input to the right by 4 bits, do it again
            until input becomes 0.    
    char[] map = {'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";
        String result = "";
        while(num != 0){
            result = map[(num & 15)] + result; 
            num = (num >>> 4);
        }
        return result;
    }

https://discuss.leetcode.com/topic/60414/java-solution

  public String toHex(int dec) {
      if (dec == 0) return "0";
      StringBuilder res = new StringBuilder();
      
      while (dec != 0) {
          int digit = dec & 0xf;
          res.append(digit < 10 ? (char)(digit + '0') : (char)(digit - 10 + 'a'));
          dec >>>= 4;
      }
      
    return res.reverse().toString();
  }

https://discuss.leetcode.com/topic/60365/simple-java-solution-with-comment/3

StringBuilder may be used here.

    public String toHex(int num) {
        StringBuilder sb = new StringBuilder();
        do {
            int n = num & 0xf;
            n += n < 0xa ? '0' : 'a' - 10;
            sb.append((char)n);
        } while ((num >>>= 4) != 0); 
        return sb.reverse().toString();
    }

X. DFS
https://discuss.leetcode.com/topic/60419/java-solution-num-0xffffffffl-to-long

    public String toHex(int num) {
        return num == 0 ? "0" : toHex(num & 0xffffffffL);
    }
    
    public String toHex(long num) {
        return num < 16 ? hexdigit(num) : toHex(num / 16) + hexdigit(num % 16);
    }
    
    private String hexdigit(long num) {
        assert num < 16;
        return num < 10 ? Character.toString((char)(num + '0')) : Character.toString((char)(num - 10 + 'a'));
    }