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.
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);
}
