https://leetcode.com/problems/divide-two-integers/description/

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

If it is overflow, return MAX_INT.

https://leetcode.com/problems/divide-two-integers/discuss/13397/Clean-Java-solution-with-some-comment.

public int divide(int dividend, int divisor) {
    //Reduce the problem to positive long integer to make it easier.
    //Use long to avoid integer overflow cases.
    int sign = 1;
    if ((dividend > 0 && divisor < 0) || (dividend < 0 && divisor > 0))
        sign = -1;
    long ldividend = Math.abs((long) dividend);
    long ldivisor = Math.abs((long) divisor);
    
    //Take care the edge cases.
    if (ldivisor == 0) return Integer.MAX_VALUE;
    if ((ldividend == 0) || (ldividend < ldivisor)) return 0;
    
    long lans = ldivide(ldividend, ldivisor);
    
    int ans;
    if (lans > Integer.MAX_VALUE){ //Handle overflow.
        ans = (sign == 1)? Integer.MAX_VALUE : Integer.MIN_VALUE;
    } else {
        ans = (int) (sign * lans);
    }
    return ans;
}

private long ldivide(long ldividend, long ldivisor) {
    // Recursion exit condition
    if (ldividend < ldivisor) return 0;
    
    //  Find the largest multiple so that (divisor * multiple <= dividend), 
    //  whereas we are moving with stride 1, 2, 4, 8, 16...2^n for performance reason.
    //  Think this as a binary search.
    long sum = ldivisor;
    long multiple = 1;
    while ((sum+sum) <= ldividend) {
        sum += sum;
        multiple += multiple;
    }
    //Look for additional value for the multiple from the reminder (dividend - sum) recursively.
    return multiple + ldivide(ldividend - sum, ldivisor);
}