convertaNumbertoHexadecimal*

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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.

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/*
要将一个十进制数转换为十六进制数，不管其是正数还是负数，都只需要将其二进制表示每四位分成一个单元，将其取出后计算这四位二进制数代表的十进制数，与0~f之间的数字做一个映射即可。要把每四位取出也很简单，与0xf进行AND运算即可。
在C++中，左移是逻辑移位，也就是说在数字后面补0，右移运算符是算术移位，也就是在左侧补符号位(正数补0，负数补1)
*/
/*
如果是负数，就直接强制转换为无符号数，直接补码变为无符号数。

正数的时候就是一个简单的十六进制转换问题。

*/
class Solution {
public:
	string toHex(int num) {
		/*string res;
		vector<char>re{ '0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f' };
		int low = 0, high = 0;
		if (num == 0) {
			return "0";
		}
		while (num&&res.size() < 8) {
			res = re[(num & 0xf)] + res;
			num = num >> 4;
		}
		return res;*/

		if (num == 0) return "0";
		string ans;
		unsigned int n;
		n = (unsigned int)num;

		while (n > 0) {
			int a = n % 16;
			string c;
			n /= 16;
			if (a > 9)
				c += 'a' + a - 10;
			else
				c += '0' + a;
			ans.insert(0, c);
		}
		return ans;
	}
};