007 easy

4/29/2015

Meet a very serious problem: overflow when reversed.

As the spoilers say, return 0 when the reversed number overflows.

Solution: 1) use LONG type to store the overflowed result. 2) add judgement of overflow or not before output it

Algorithm 1

(Came by my first thought)

Meet another problem of dealing with overflow number. The problem came with j.

Problem: j is the number 1,10,100… 1,000,000,000. When dealing with the first digit of 3(like 1 when x=1,000,000,003) j=1,000,000,000 from the last loop. but in this loop, sum will be added to 3,000,000,000, theoretically. However, because j overflows as an integer, the result sum in this loop messed up.

Two solutions:

S1: set j as long type

S2: sum=(long ) array[i]j+sum; OR sum=array[i](long) j+sum; use forced type transfer

package happynumber;public class ReverseInteger {    public static int reverse (int x) {        int i;i=0;        //long type is for compare. Only larger range can compare Max and Min Integer                long sum=0;        int[] array=new int[33];        int number;        number=x;        if(x<0) number=-x;        System.out.println(number);        System.out.println(number);        while(number!=0) {            array[i]=number%10;            System.out.println(array[i]);            number=number/10;            i++;        }        long j;j=1;             i=i-1;        for(;i>=0;i--){            sum=array[i]*j+sum;            System.out.println(sum);            j=j*10;            System.out.println(j);        }        if(x<0) sum=-sum;        System.out.println("original result");        System.out.println(sum);        if(sum>Integer.MAX_VALUE || sum<Integer.MIN_VALUE) {            System.out.println("overflow");            return 0;        }        else {            System.out.println((int)sum);            return (int) sum;        }    }    public static void main(String args[]) {        int y=0;        y=reverse(-1000000003);    }}

Algorithm 2: change about calculating result algorithm

package happynumber;public class ReverseInteger2 {        public static int reverse (int x) {            long result,number;            result=0;            number=x;            if(x<0) number=-number;            while(number!=0) {                result=result*10+number%10;                number=number/10;            }            if(x<0) result=-result;            System.out.println("original result");            System.out.println(result);            if(result>Integer.MAX_VALUE || result<Integer.MIN_VALUE){                System.out.println(0);                return 0;            }            else {                System.out.println((int)result);                return (int) result;            }        }        public static void main(String args[]) {            int y=0;            y=reverse(1000000003);        }}