1.题目

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.
给一个32位整数，求16进制表示。负数用补码。

2.分析

一开始挺傻的，老老实实按16进制转换来做，负数还求了反码+1.
然而！！错了，因为-2147483648求反码的时候会溢出。因为正数最大表示为2147483647。不过这个问题后来也可以通过将int转换为long long来解决。

偷瞄了solution，原来别人都是用位运算来做的。唉，刷了这么久的题也没学会套路。
就想字符串一类的题，每次都老老实实遍历if else一大堆，到头来一看solution，人家一行正则表达式就搞定了。还是得长点心啊。

3.代码

正解。别想着自己转换进制了，数在计算机里本来就是按二进制存的。多用位运算。

string toHex(int n) {
    if (n == 0)
        return "0";
    string HEX = "0123456789abcdef";
    vector<char> result;
    int count = 0;
    while (n && count++ < 8) {
        result.push_back(HEX[n & 0xf]);
        n >>= 4;
    }
    return string(result.rbegin(), result.rend());

我的挫挫的解

class Solution {
public:
    string toHex(int n) {
        long long num=n;
        if (num == 0)
            return "0";
        vector<char> hex;
        vector<int> neg;
        string match= "0123456789abcdef";
        bool sign = num > 0 ? true : false;
        num = num > 0 ? num : -num;
        while (num > 0) {
            if(sign)//正数，原码
                hex.push_back(match[num % 16]);
            else//负数，反码
                neg.push_back(15-num % 16);
            num /= 16;
        }
        if (!sign)//反码+1
        {
            for (int i = neg.size(); i < 8; i++) {
                neg.push_back(15);
            }
            int carry = 1; 
            for (int i = 0; i < neg.size(); i++) {
                carry = neg[i] + carry;
                hex.push_back(match[carry % 16]);
                carry = carry / 16;
            }
        }
        return string(hex.rbegin(), hex.rend());
    }
};