[LeetCode] 29. Divide Two Integers

最新推荐文章于 2019-11-06 15:25:17 发布

Peng_maple

最新推荐文章于 2019-11-06 15:25:17 发布

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分类专栏： leetcode 200

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本文介绍了一种不使用乘法、除法和取余操作来实现两个整数相除的方法。通过位运算技巧，实现了整数除法的功能，并考虑了32位有符号整数范围内的溢出情况。

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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.

解析

用位运算

class Solution {
public:
    int divide(int num1, int num2){
        long long dividend = num1 < 0 ? -(long long)num1 : num1;
        long long divisor = num2 < 0 ? -(long long)num2 : num2;

        long long result = 0;
        while(dividend >= divisor){
            long long tmp = divisor;
            for(int i=0;dividend>=tmp;i++,tmp <<= 1){
                dividend -= tmp;
                result += (1<<i);
            }
        }
        if((num1^num2) <0)
            result = 0-result;
        if(result > INT_MAX || result < INT_MIN)
            return INT_MAX;
        return int(result); 
    }
    
};