Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
Solution:
int reverse(int x) {
long int isNegative = x<0?-1:1;
long int reversed = 0;
long num = x * isNegative;
while(num) {
reversed *= 10;
reversed += num%10;
num/=10;
}
if((isNegative == -1 && reversed > 2147483648) || (isNegative == 1 && reversed > 2147483647 )) {
return 0;
}
int ret = (int) reversed * (int)isNegative;
return ret;
}
这个题目和今日 lab 里面的 itoa 比较类似。但是由于 reverse integer 只同样是输出 integer，因此可以把 reversed * 10 后，直接将得到的数字加在reversed后面。
附 itoa：
char* itoa(int num, char result[]) {
long int num2 = num;
if(num == 0) {
result[0] = '0';
} else {
if(num<0){
num2 = - (long int)num;
result[0] = '-';
long int base = 1;
int digit = 1;
while(base<=num2){
base *= 10;
digit++;
}
result[digit--] = '\0';
while(num2!=0){
int last = num2 % 10;
char temp = last + '0';
result[digit--] = temp;
num2 /= 10;
}
} else {
long int base = 1;
int digit = 1;
while(base<=num){
base *= 10;
digit++;
}
digit--;
result[digit --] = '\0';
while(num2!=0){
int last = num2 % 10;
char temp = last + '0';
result[digit--] = temp;
num2 /= 10;
}
}
}
return result;
}
