Reverse Integer - LeetCode

Problem:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321


public class Solution {
    public int reverse(int x) {
    }
}

public class Solution {

public int reverse(int x) {

}

Have you thought about this?
1 Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
2. If the integer’s last digit is 0, what should the output be? ie, cases such as 10, 100.
3. 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?
4. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Solution


public class Solution {
    public int reverse(int x) {
    	int sign = x < 0 ? -1 : 1;
    	x = Math.abs(x);
    	
    	long val = 0;
    	while(x != 0) {
    		int mod = x%10;
    		val = val * 10 + mod;
    		x = x / 10;
    	}
    	val *= sign;
    	if(val > Integer.MAX_VALUE || val < Integer.MIN_VALUE) {
    		val = 0;
    	}
    	
    	return (int)val;
    }
}

public class Solution {

public int reverse(int x) {

int sign = x < 0 ? -1 : 1;

x = Math.abs(x);

long val = 0;

while(x != 0) {

int mod = x%10;

val = val * 10 + mod;

x = x / 10;

}

val *= sign;

if(val > Integer.MAX_VALUE || val < Integer.MIN_VALUE) {

val = 0;

}

return (int)val;

}

Another solution (a little bit slower)


    public static int reverse(int x) {
    	boolean positive = x >= 0 ? true : false;
    	StringBuilder tmp = new StringBuilder(String.valueOf(x).replace("-", ""));
    	if(!positive) {
    		tmp.reverse().insert(0, '-');
    	} else {
    		tmp.reverse();
    	}
    	
    	long val = Long.valueOf(tmp.toString());
    	if(val > Integer.MAX_VALUE || val < Integer.MIN_VALUE) {
    		val = 0;
    	}
   		return (int)val;
    }