29. Divide Two Integers

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

Description

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: $[-2^{31},2^{31}-1]$ . For the purpose of this problem, assume that your function returns $2^{31}-1$ when the division result overflows.

Solution

class Solution {
public:
    int divide(int dividend, int divisor) {
        if(dividend==divisor){
            return 1;
        }
        // divident=-2^31 , divisor=-1时，相除会越界
        if(!(dividend^0x80000000) && (divisor==-1)){
            return 0x7fffffff;
        }
        // divisor=-2^31 作为除数，并且 divident与其不相等
        // 返回必为 0
        if(!(divisor^0x80000000)){
            return 0;
        }
        int sign = 1;
        int ret = 0;
        // 确认除数与被除数是否异号
        if((dividend&0x80000000)^(divisor&0x80000000)){
            sign = -1;
        }
        // 相当于 divisor = abs(divisor)
        if(divisor&0x80000000){
            divisor = (~divisor)+1;
        }
        if(!(dividend^0x80000000)){
            ret += 1;
            dividend += divisor;
        }
        // 相当于 dividend = abs(dividend)
        if(dividend&0x80000000){
            dividend = (~dividend)+1;
        }
        int i = 31;
        // 获取除数的最高的有效位
        while(!(divisor & (1<<i))){
            --i;
        }
        // 使用类似辗转相除法
        for(int j=30-i;j>=0;--j){
            while((divisor<<j)<=dividend){
                dividend -= (divisor<<j);
                ret += (1<<j);
            }
        }
        if(sign==-1){
            // 取相反数：ret = -ret
            ret = (~ret)+1;
        }
        return ret;
    }
};