Question

29. Divide Two Integers

Solution

Main idea of the title: Given two numbers, it is required that multiplication, division and redundancy operations should not be used to derive their quotients.

Ideas: Let's talk about the method of using displacement to realize it.

7/3 = 2, 7 is dividend, 3 is divisor
 Let's move the divisor to the left, assuming that it moves n times and moves to the nearest dividend, when the divisor = dividend-divisor, part of quotient is 2^n.
If the dividend > the dividend, the loop continues
 When the divisor moves to the left and moves m times, it moves to the nearest dividend, where the divisor = dividend-divisor and the quotient part is 2^m.
The final quotient is 2^n+2^m +...

Java implementation:

Method 1: If divisions can be used, one step will do.

public int divide2(int dividend, int divisor) {    
    // overflows
    if (dividend == Integer.MIN_VALUE && divisor == -1) return Integer.MAX_VALUE;
    // Given two numbers, finding their quotients requires that multiplication, division and redundancy operations should not be used.
    return dividend / divisor;
}

Law 2: The following is achieved by subtraction, execution timeout

public int divide2(int dividend, int divisor) {
    // overflows
    if (dividend == Integer.MIN_VALUE && divisor == -1) return Integer.MAX_VALUE;

    int ans = 0;
    boolean negative = !((dividend > 0 && divisor > 0) || (dividend < 0 && divisor < 0));
    dividend = Math.abs(dividend);
    divisor = Math.abs(divisor);
    dividend -= divisor;
    while (dividend >= 0) {
        ans++;
        dividend -= divisor;
    }
    return negative ? -ans : ans;
}

Method 3: Implemented by Shift

public int divide(int dividend, int divisor) {
    // Prevent overflow
    if (dividend == Integer.MIN_VALUE && divisor == -1) return Integer.MAX_VALUE;

    // Symbols to get the final result
    int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
    long dvd = Math.abs((long) dividend);
    long dvs = Math.abs((long) divisor);
    int ans = 0;
    while (dvd >= dvs) {
        long tmp = dvs, multiple = 1;
        while (dvd >= (tmp << 1)) {
            tmp <<= 1;
            multiple <<= 1;
        }
        dvd -= tmp;
        ans += multiple;
    }
    return sign == 1 ? ans : -ans;
}