[2 ms JAVA] Better way to handle overflow/underflow and a simple solution

  • 1

    This solution would work if you have been allowed to use 64 bits instead of 32 bit

    public class Solution {
    static int max = Integer.MAX_VALUE;
    static int min = Integer.MIN_VALUE;
    public int reverse(int x) {
        int r = 0;
        if(x>1000000000 && x%10>2) return 0; // optimization only for 32 bit
        if(x< -1000000000 && x%10<-2) return 0; // optimization only for 32 bit
        while(x != 0)
            if((r*10)/10 != r) return 0; // check overflow without long
            r = r*10 + x%10;
            x = x/10;
        return r;

  • 0

    could you explain the two statements given by you for optimization? thanks!

  • 0

    This optimization is nothing but a fast way to check if the reverse of the given value will underflow/overflow.

    How it works?

    Let's take a basic example:
    x = 1000000003
    it would satisfy x>1000000000
    and when reversed it would overflow (3000000001)

    We can check most of the cases if they over/underflow without going into the loop
    Mind you that this would only work for 32 bit integers, for long you might have to remove/change this optimization.

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.