*Divide Two Integers

標籤：

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

If it is overflow, return MAX_INT.

 

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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);}

reference: https://leetcode.com/discuss/57346/clean-java-solution-with-some-comment

*Divide Two Integers