29. Divide Two Integers

Posted 2021-01-11 ruruozhenhao

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.

AC code:

class Solution {
public:
    typedef long long ll;
    int divide(int dividend, int divisor) {
        if (dividend == INT_MIN && divisor == -1) return INT_MAX;
        int negative_nums = 0;
        ll num, ans;
        int res=0;
        if (dividend < 0) 
            negative_nums++;
        if (divisor < 0) 
            negative_nums++;
        ll l_dividend = abs((ll)dividend);
        ll l_divisor = abs((ll)divisor);
        while (l_dividend >= l_divisor) {
            num = l_divisor;
            ans = 1;
            while (1) {
                num = num << 1;
                if (l_dividend >= num) {
                    ans <<= 1;
                } else 
                    break;
            }        
            num = num>>1;
            l_dividend -= num;
            res += ans;
        }
        if (negative_nums%2 == 0)
            return res;
        else 
            return -res;
    }
};

Runtime: 12 ms, faster than 98.72% of C++ online submissions for Divide Two Integers.