题意:将一个十进制数转换为16进制数。
解题思路:
方法一:常规方法,对该十进制数判断正负,正数就是十进制数直接转化为十六进制数,负数就是先变为正数然后求补码转换为十六进制数。
该方法比较麻烦。
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:对该十进制数每四位对十六进制数的f求与,这样可以得到该数的低4位数的值,然后将该值转换为十六进制数,然后将该数右移4位。由于对于正数,右移时最高位补0,对于负数右移时最高位补1,所以要判断右移的次数,最多不能超过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;
}
};