2119. A Number After a Double Reversal

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problem

solution

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class Solution {
public:
    int reverse(int num){
        int ret = 0;
        while(num){
            ret*=10;
            ret+=(num%10);
            num/=10;
        }
        cout<<ret<<endl;
        return ret;
    }
    bool isSameAfterReversals(int num) {
        return reverse(reverse(num)) == num;
        
    }
};

analyze

  • time complexity O(1)
  • space complexity O(1)