【LeetCode】405 Convert a Number to Hexadecimal (java实现)

来源:互联网 时间:2016-11-10

原题链接

https://leetcode.com/problems/convert-a-number-to-hexadecimal/

原题

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"

题目要求

题目叫“将数字转化为十六进制”,顾名思义,这里需要注意的是数字是包含负数的,所以如果方法不很合适,处理起来会稍微麻烦一些。

要求:

  1. 转化后的十六进制字符串都是小写的;
  2. 十六进制字符串不能以0开头(如果只有一个0除外);
  3. 数字大小在32bit范围内,不用担心处理数据时溢出;
  4. 不能使用库里的转化和格式打印;

    解法

    解法一:最原始的方法,完全按照数字的源码、反码、补码的格式来转化,这种思路下,就要先将数字转化为2进制,再将二进制转化为十六进制。同时,还需要注意数字为负数时,需要一些特殊的操作。这种解法非常麻烦,但是却非常直接。

    public String toHex(int num) {

    if (num == 0) {

    return "0";

    }

    int MAX = 32;

    boolean isNegative = false;

    int bits[] = new int[MAX];

    if (num < 0) {

    isNegative = true;

    bits[MAX - 1] = 1;

    num = -num;

    }

    int i = 0;

    // 转化为二进制的原码

    while (num > 0) {

    bits[i++] = num % 2;

    num /= 2;

    }

    // 如果是负数,需要取反并且+1从而得到补码

    if (isNegative) {

    // 取反

    for (int j = 0; j < bits.length - 1; j++) {

    bits[j] = (bits[j] + 1) % 2;

    }

    // +1

    int digit = 1;

    int res = 0;

    for (int j = 0; j < bits.length - 1; j++) {

    res = bits[j] + digit;

    bits[j] = res % 2;

    digit = res / 2;

    }

    }

    // 二进制转化为十六进制

    String ret = "";

    for (int j = 0; j < bits.length; j += 4) {

    int data = 0;

    for (int j2 = 0; j2 < 4; j2++) {

    data += bits[j + j2] * (1 << j2);

    }

    ret = String.format("%x", data) + ret;

    }

    // 去掉字符串前面多余的0

    for (int j = 0; j < ret.length(); j++) {

    if (ret.charAt(j) != '0') {

    ret = ret.substring(j);

    break;

    }

    }

    return ret;

    }

    解法二:第二种解法就是按位与来获取。既然是得到十六进制,那么每次与上0xF(二进制就是1111),得到一个值,然后数字向右移动4位,这里需要注意的是数字是有符号的,刚好可以利用Java提供的无符号移动>>>。完美!!!

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 & 0xF)] + result;

num = (num >>> 4);

}

return result;

}

相关阅读:
Top