29. Divide Two Integers

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Problem

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

The integer division should truncate toward zero, which means losing its fractional part. For example, 8.345 would be truncated to 8, and -2.7335 would be truncated to -2.

Return the quotient after dividing dividend by divisor.

Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: $[−2^{31}, 2^{31} − 1]$. For this problem, if the quotient is strictly greater than $2^{31} - 1$, then return $2^{31} - 1$, and if the quotient is strictly less than $-2^{31}$, then return $-2^{31}$.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = 3.33333.. which is truncated to 3.

Example 2:

Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = -2.33333.. which is truncated to -2.

Constraints:

• $-2^{31} <= dividend, divisor <= 2^{31} - 1$
• divisor != 0

Code

class Solution {
    public int divide(int dividend, int divisor) {
        int sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1;
        
        double res = 0;
        double dd = Math.abs((double)dividend);
        double dv = Math.abs((double)divisor);
        
        while(dd - dv >= 0) {
            double pow = getPow(dd, dv);
            dd -= pow * dv;
            res += pow;
        }
        
        res *= sign;
        if(res >= Integer.MAX_VALUE){
            return Integer.MAX_VALUE;
        } else {
            return (int)res;
        }
    }
    
    private double getPow(double dd, double dv) {
        double temp = 0;
        while(dd - Math.pow(2.0, temp) * dv >= 0) {
            temp++;
        }
        
        return Math.pow(2.0, temp - 1);
    }
}

class Solution {
    public int divide(int dividend, int divisor) {
        // 判断符号的方法
        int sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1;

        long dd = Math.abs((long)dividend);
        long dv = Math.abs((long)divisor);

        long res = 0;

        while(dv <= dd){
            long temp = dv;
            long mul = 1;
            while(dd >= (temp << 1)){
                temp <<= 1;
                mul <<= 1;
            }

            res += mul;
            dd -= temp;
        }

        res *= sign;

        if(res >= Integer.MAX_VALUE){
            return Integer.MAX_VALUE;
        } else {
            return (int)res;
        }
    }
}

Complexity

时间复杂度：O(log(n)) 空间复杂度：O(1)