Reverse Integer

Question

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

Answer 1

int reverse(int x) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
    int lastDigit = 0;
    int result = 0;
    bool isNeg = x>0? false:true;
    x = abs(x);

    while(x>0)
    {
        lastDigit = x%10;
        result = result*10 + lastDigit;
        x = x/10;
    }

    if(result<0) return -1;

    if(isNeg)
        result *=-1;

    return result;        
}

Answer 2

public int reverse(int x){

    // Start typing your Java solution below
    // DO NOT write main() function

    int r = 0;

    while(x != 0) {
        r = r*10 + x % 10;
        x /= 10;
    }
    return r;
}

Answer 3

class Solution {
public:
int reverse(int x) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
        bool zheng = (x > 0) ? true : false;
int temp = abs(x);
int res = 0;
while (temp > 0)
{
    res = res*10 + temp%10;
    temp /= 10;
}
if(zheng)
    return res;
else
    return -1*res;
}
};

Answer 4

class Solution { public: int reverse(int x) { // Start typing your C/C++ solution below // DO NOT write int main() function

    int absx;
    int modx=1, leftx=11; 
    int sum=0;

    if(x<0) absx=-1*x; 
    if(x>0) absx=x; 
    if(x==0) return x;

    while(modx>0){
        modx=absx/10; 
        leftx=absx-10*modx;
        absx=modx; 
        sum=sum*10+leftx; }

    if(x<0) return -sum; 
    else return sum;

}

};

Answer 5

class Solution { public:

int reverse(int x)

{ int r=0;

while(x!=0) {

r=r*10+x%10; x=x/10; } return r; } };

Answer 6

//my code shows the history how to reduce LOC
int reverse(int x) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
    int result =0;
    //int remainder;
    //bool boolX = (x>0?true:false);

    //if(boolX == false) x = -1 *x;

    while(x !=0 ){
        //cout << "x: " << x << endl; 
        result = result *10 + x % 10;
        //remainder = x % 10;
        //result += x % 10;
        x = x /10;

        //cout << "result: " << result << endl;

    }

    //result += x;

    //if(boolX == false) return -1 *result; 
    return result;

}

Answer 7

public int reverse(int x) {
    if(x==0)
        return 0;

    StringBuilder sb = new StringBuilder();

    if(x<0)
        sb.append("-");

    x=Math.abs(x);
    while(x>0){
        sb.append(x%10);
        x=x/10;
    }

    return Integer.parseInt(sb.toString());
}

Answer 8

int reverse(int x) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
    stack<int> s;
    int p = 1;

    if (x < 0) {
        p = -1;
        x = x * p;
    }

    while (x > 0) {
        int digit = x % 10;
        s.push(digit);
        x = x / 10;
    }

    int r = 0, i = 0;
    while (!s.empty()) {
        int digit = s.top();
        s.pop();
        r += digit * pow(10, i);
        i++;
    }

    return r*p;
}

Answer 9

public class Solution {

public int reverse(int x) {
    int neg = (x<0)?-1:1;
    x *= neg;

    int cnt = 10, rev = 0;
    while(x>=cnt/10){
        rev = rev*10 + (x%cnt - x%(cnt/10)) / (cnt/10);
        cnt *= 10;
    }

    return rev*neg;
}

}

Answer 10

        int reverse(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function

        int rem = x;
        int result = 0;

        while (rem != 0)
        {
            result = result * 10 + rem % 10;
            rem /= 10;            
        }

        return result;
    }

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