29. Divide Two Integers (Medium)

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

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3

Example 2:

Input: dividend = 7, divisor = -3
Output: -2

Note:

  • Both dividend and divisor will be 32-bit signed integers.
  • The divisor will never be 0.
  • Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 231 − 1 when the division result overflows.

Solutions

class Solution {
    /*
        * This problem is really frustrating with a mass of corn cases.
        * Interviewers much take good care of overflow issue. To get rid of overflow,
        * It's a good habit to convert all the variables to long type before calculation.
    */
    public int divide(int dividend, int divisor) {
        // Be careful of the overflow issue
        if (dividend == Integer.MIN_VALUE && divisor == -1) {
            return Integer.MAX_VALUE;
        }

        if (divisor == 1) {
            return dividend;
        }

        if (divisor == -1) {
            return -dividend;
        }

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

        // do not use below statements, Math.abs can be handle int overflow issue
//            long tmpDvd = Math.abs(dividend);
//            long tmpDvs = Math.abs(divisor);

        long tmpDvd = dividend;
        long tmpDvs = divisor;

        tmpDvd = Math.abs(tmpDvd);
        tmpDvs = Math.abs(tmpDvs);

        int ans = 0;

        long remain = tmpDvd;
        while (true) {
            if (remain - tmpDvs < 0) {
                break;
            }

            int count = 0;
            // 1. comparision must be >= 0, not > 0
            // 2. tmpDvs must be long type, otherwise overflow takes place
            while (remain - (tmpDvs << count) >= 0) {
                count++;
            }

            count--;

            // to reach here, count must be >= 0

            ans += (1 << count);

            // notice remain - divisor << count is absolutely different from remain - (divisor << count)
            remain -= tmpDvs << count;
        }

        return isNegative ? -ans : ans;
    }
}

Incorrect Solutions

References

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