# 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:26Output:``"1a"``

Example 2:

``Input:-1Output:``"ffffffff"``

# 题目要求

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;}// +1int 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0for (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;``}``

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