Leetcode Reverse Integer problem solution YASH PAL, 31 July 2024 In this Leetcode Reverse Integer problem solution we have given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 – 1], then return 0. Problem solution in Python. def reverse(x): sign = -1 if x<0 else 1 res, x = 0, abs(x) while x: res = res*10 + (x%10) x /= 10 # handle the overflow bound if res > 2**31+1 or res < -2**31-1: return 0 return res*sign Problem solution in Java. public int reverse(int x) { boolean neg = false; if(x<0){ neg = true; x = -x; } long ans = 0; int maxPow = (int)Math.log10(x); while(x>0){ ans+= (x%10 * Math.pow(10,maxPow--)); x=x/10; } if(ans > Integer.MAX_VALUE){ return 0; } if(neg){ return (int)(-ans); }else{ return (int) ans; } } Problem solution in C++. public: int reverse(int x) { long res = 0; while (x != 0) { res = 10 * res + x % 10; x /= 10; } return (res > INT_MAX || res < INT_MIN) ? 0 : res; } }; Problem solution in C. int reverse(int x) { long reval = 0; for(;x;x/=10) reval = reval *10 + x%10; x = reval; if(reval != x) return 0; return reval; } coding problems