題意：將一個(gè)十進(jìn)制數(shù)轉(zhuǎn)換為16進(jìn)制數(shù)。
解題思路：
方法一：常規(guī)方法，對(duì)該十進(jìn)制數(shù)判斷正負(fù)，正數(shù)就是十進(jìn)制數(shù)直接轉(zhuǎn)化為十六進(jìn)制數(shù)，負(fù)數(shù)就是先變?yōu)檎龜?shù)然后求補(bǔ)碼轉(zhuǎn)換為十六進(jìn)制數(shù)。
該方法比較麻煩。

class Solution {
public:
    string toHex(int num) {
        if(num == 0x80000000) return "80000000";
        if(num == 0) return "0";
        int flg = 0;
        if(num < 0) flg = -1;
        num = abs(num);
        string h_num;
        while(num){
            int m = num % 16;
            if(m < 10)
                h_num = char(m + '0') + h_num;
            else
                h_num = char(m - 10 + 'a') + h_num;
            num /= 16;
        }
        if(flg == -1){
            int i;
            for(i = 0; i < h_num.size(); i++){
                int m;
                if('0' <= h_num[i] && h_num[i] <= '9')
                    m = 15 - (h_num[i] - '0');
                else
                    m = 15 - (10 + (h_num[i] - 'a'));
                if(m < 10)
                    h_num[i] = char(m + '0');
                else
                    h_num[i] = char(m - 10 + 'a');
            }
            while(i < 8){
                h_num = "f" + h_num;
                i++;
            }
            h_num[7]++;
            if(h_num[7] == '9' + 1)
                h_num[7] = 'a';
            else if(h_num[7] == 'g'){
                int tmp = 1, i = 7;
                while(tmp){
                    h_num[i--] = '0';
                    h_num[i]++;
                    if(h_num[i] == '9' + 1)
                        h_num[i] = 'a';
                    else if(h_num[i] == 'g')
                        continue;
                    tmp = 0;
                }
            }
        }
        return h_num;
    }
};

方法二：對(duì)該十進(jìn)制數(shù)每四位對(duì)十六進(jìn)制數(shù)的f求與，這樣可以得到該數(shù)的低4位數(shù)的值，然后將該值轉(zhuǎn)換為十六進(jìn)制數(shù)，然后將該數(shù)右移4位。由于對(duì)于正數(shù)，右移時(shí)最高位補(bǔ)0，對(duì)于負(fù)數(shù)右移時(shí)最高位補(bǔ)1，所以要判斷右移的次數(shù)，最多不能超過(guò)8次。

class Solution {
public:
    string toHex(int num) {
        if(num == 0) return "0";
        string HEX = "0123456789abcdef";
        string ans;
        int count = 0;
        while(num && count++ < 8){
            ans = HEX[num & 0xf] + ans;
            num >>= 4;
        }
        return ans;
    }
};