ConvertANumberToHexadecimalOPT [source code]

public class ConvertANumberToHexadecimalOPT {
static

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

    public static void main(String[] args) {
        ConvertANumberToHexadecimalOPT.Solution tester = new ConvertANumberToHexadecimalOPT.Solution();
        int input1 = -100001;
        System.out.println(Integer.toHexString(input1));
        System.out.println(tester.toHex(input1));
    }
}

这个是 subission 最优解:7(83).

思路还是很直接的, 就是用bit shift来做, 很聪明, 一个很简单的循环就解决了. 我确实是把问题搞复杂了, 而且对于bit manipulation 我现在确实还是有点不够熟练.

这个题目展示的一个bit技巧就是用&F这种类似的操作来提取指定位置的 bit. 这个其实以前是见过的. 一旦开窍看懂这个技巧, 这个问题的思路就简单很多了;

这个是用 StringBuilder 的版本:

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();  
}

Problem Description

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.
Example 1:

Input:
26

Output:
"1a"
Example 2:

Input:
-1

Output:
"ffffffff"
Difficulty:Easy
Category:Algorithms
Acceptance:40.99%
Contributor: LeetCode
Related Topics
bit manipulation