Reverse Integer - Handling overflow

  • 0
    class Solution {
        public int reverse(int x) {
            int reverseNumber = 0;
            int prevReverse =0;
            while(x != 0){
                prevReverse = reverseNumber;
                reverseNumber = (reverseNumber*10) + (x%10);
                x = x/10;
                if(reverseNumber/10 != prevReverse){
                    return 0;
            return reverseNumber;

    if(reverseNumber/10 != prevReverse){
    If the number is too large for datatype it can cause overflow and return a garbage value. The idea is to compare with the previous number to identify if there was an overflow at this point. If previous number and (reverseNumber/10) doesnt match return zero instead of garbage value.

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