Integer.MIN_VALUE除以-1

Question

why this line of code matters?(I get a wrong answer without it)

if (dividend == Integer.MIN_VALUE && divisor == -1) {
    return Integer.MAX_VALUE;
}

The question:

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return 2147483647

Answer

public int divide(int dividend, int divisor) {

    if(divisor == 0){
        return dividend > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
    }

    if(dividend == 0){
        return 0;
    }

    if (dividend == Integer.MIN_VALUE && divisor == -1) {
        return Integer.MAX_VALUE;
    }

    boolean isNeg = (dividend > 0 && divisor < 0) || (dividend < 0 && divisor > 0);


    Long up = Math.abs((long) dividend);
    Long down = Math.abs((long) divisor);

    int res = 0;

    while(up >= down){
        int shift = 0;

        while(up >= (down << shift)){
            shift++;
        }

        up -= down << (shift - 1);
        res += 1 << (shift - 1);
    }

    return isNeg ? -res : res;
}

Answer 1

Because, the absolute values of Integer.MAX_VALUE and Integer.MIN_VALUE are not equal.

• Integer.MAX_VALUE is 2147483647
• Integer.MIN_VALUE is -2147483648

In case you divide Integer.MIN_VALUE by -1 , the value would overflow ( 2147483648 > 2147483647 ), thus there must be a limit for this operation.

Answer 2

Java use 32 bit to store int .

The max int value is 2 ³¹ -1

0111 1111 1111 1111 1111 1111 1111 1111

The min int value is -2 ³¹

1000 0000 0000 0000 0000 0000 0000 0000

In other words, int does not have a big enough value to store 2 ³¹ ( -Integer.MIN_VALUE ).