Problem

Given an integer num, return a string representing its hexadecimal representation. For negative integers, two’s complement method is used.

All the letters in the answer string should be lowercase characters, and there should not be any leading zeros in the answer except for the zero itself.

Note: You are not allowed to use any built-in library method to directly solve this problem.

Examples

Example 1:

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Input:
num = 26
Output:
 "1a"

Example 2:

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Input:
num = -1
Output:
 "ffffffff"

Solution

Method 1 - Iterative

The idea is that each time we take a look at the last four digits of binary version 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.

Code

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class Solution:
    def toHex(self, num: int) -> str:
        if num == 0:
            return '0'
        chars = '0123456789abcdef'
        s = []
        for i in range(7, -1, -1):
            x = (num >> (4 * i)) & 0xF
            if s or x != 0:
                s.append(chars[x])
        return ''.join(s)
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class Solution {
    public String toHex(int num) {
        if (num == 0) {
            return "0";
        }
        StringBuilder sb = new StringBuilder();
        while (num != 0) {
            int x = num & 15;
            if (x < 10) {
                sb.append(x);
            } else {
                sb.append((char) (x - 10 + 'a'));
            }
            num >>>= 4;
        }
        return sb.reverse().toString();
    }
}
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class Solution {
    public String toHex(int num) {
        if (num == 0) {
            return "0";
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 7; i >= 0; --i) {
            int x = (num >> (4 * i)) & 0xf;
            if (sb.length() > 0 || x != 0) {
                char c = x < 10 ? (char) (x + '0') : (char) (x - 10 + 'a');
                sb.append(c);
            }
        }
        return sb.toString();
    }
}
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class Solution {
public:
    string toHex(int num) {
        if (num == 0) return "0";
        string s = "";
        for (int i = 7; i >= 0; --i) {
            int x = (num >> (4 * i)) & 0xf;
            if (s.size() > 0 || x != 0) {
                char c = x < 10 ? (char) (x + '0') : (char) (x - 10 + 'a');
                s += c;
            }
        }
        return s;
    }
};
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func toHex(num int) string {
	if num == 0 {
		return "0"
	}
	sb := &strings.Builder{}
	for i := 7; i >= 0; i-- {
		x := num >> (4 * i) & 0xf
		if x > 0 || sb.Len() > 0 {
			var c byte
			if x < 10 {
				c = '0' + byte(x)
			} else {
				c = 'a' + byte(x-10)
			}
			sb.WriteByte(c)
		}
	}
	return sb.String()
}