[LeetCode]405. Convert a Number to Hexadecimal

Posted on Views: Reading time ≈ 1 mins.

题目

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:

1
2
3
4
Input:
26
Output:
"1a"

Example 2:

1
2
3
4
Input:
-1
Output:
"ffffffff"

难度

Easy

方法

对于num，每次num&15便可取到最右的4个bit位，然后num>>4循环处理即可将num转换为16进制。因为num范围为32bit位，因此最多右移8次，每次右移4个bit位。当num不为0时，转换为16进制后，前面有0需去除。

python代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution(object):
    def toHex(self, num):
        """
        :type num: int
        :rtype: str
        """
        if num == 0:
            return "0"
        map = ["0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f"]
        hex = ""
        for i in range(8):
            hex = map[num & 0xf] + hex
            num = num >> 4

        return hex.lstrip("0")

assert Solution().toHex(-1) == "ffffffff"
assert Solution().toHex(26) == "1a"