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algorithm notes: leetcode#405 convert a number to hexadecimal

admin 11月 12, 2020 0

Problem

Solution

Initial thoughts

Python implementation

class :
    def toHex(self, num):
        if 0 <= num < 10:
            return str(num)
        if num < 0:
            num = num + 2**32
        chars = [str(i) if i < 10 else chr(i-10+ord('a')) for i in range(16)]
        res = []
        while num:
            res.append(chars[num&15])
            num >>= 4
        return "".join(res[::-1])

Java implementation

class  {
    public String toHex(int num) {
        if(num >= 0 && num < 10){
            return Integer.toString(num);
        }
        char[] map = {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};
        String res = "";
        while(num != 0){
            res = map[num & 15] + res;
            num >>>= 4;
        }
        return res;
    }
}

Time complexity

O(logN)

Space complexity

O(1)

Links

405. Convert a Number to Hexadecimal
(中文版) 算法笔记: 力扣#405 数字转换为十六进制数