Leetcode 29 :Divide two integers

整数除法算法

最新推荐文章于 2020-03-20 22:34:42 发布

原创最新推荐文章于 2020-03-20 22:34:42 发布 · 161 阅读

0 ·

CC 4.0 BY-SA版权

oj数学题专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种不使用乘法、除法和取模运算符实现整数除法的算法。通过逐步增加除数的方式减少计算复杂度，并考虑了各种边界情况以避免溢出。

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.

class Solution {
public:
    int divide(int dividend, int divisor) {
	if (divisor == 0 || divisor == -1 && dividend == INT_MIN) // 特殊情况会导致结果溢出
		return INT_MAX;
	int result = 0;
	long long int Divisor = divisor;
	long long int Dividend = dividend;
	if (Dividend > 0 && Divisor > 0)
	{
		if (Dividend >= Divisor * 2)
		{
			result = 1;
			Divisor = Divisor * 2; // 采用一个加倍扩大除数的方法，可以将时间复杂度变成O(logn+n/2),n为输入被除数的大小
			while (Dividend >= Divisor)
			{
//				Dividend = Dividend - Divisor;
				Divisor = Divisor * 2;
				result = result * 2;
			}
		}
		Dividend = Dividend - result * divisor;
		Divisor = divisor;
		while (Dividend >= Divisor)
		{
			Dividend = Dividend - Divisor;
			result += 1;
		}
	}
	if (Dividend > 0 && Divisor < 0)
	{
		if (Dividend + Divisor * 2 >= 0)
		{
			result = -1;
			Divisor = Divisor * 2;
			while (Dividend + Divisor >= 0)
			{
				Divisor = Divisor * 2;
				result = result * 2;
			}
		}
		Dividend = Dividend - result * divisor;
		Divisor = divisor;
		while (Dividend + Divisor >= 0)
		{
			Dividend = Dividend + Divisor;
			result -= 1;
		}
	}
	if (Dividend < 0 && Divisor > 0)
	{
		if (Dividend + Divisor * 2 <= 0)
		{
			result = -1;
			Divisor = Divisor * 2;
			while (Dividend + Divisor <= 0)
			{
				//				Dividend = Dividend - Divisor;
				Divisor = Divisor * 2;
				result = result * 2;
			}
		}
		Dividend = Dividend - result * divisor;
		Divisor = divisor;
		while (Dividend + Divisor <= 0)
		{
			Dividend = Dividend + Divisor;
			result -= 1;
		}
	}
	if (Dividend < 0 && Divisor < 0)
	{
		if (Dividend <= Divisor * 2)
		{
			result = 1;
			Divisor = Divisor * 2;
			while (Dividend <= Divisor)
			{
				//				Dividend = Dividend - Divisor;
				Divisor = Divisor * 2;
				result = result * 2;
			}
		}
		Dividend = Dividend - result * divisor;
		Divisor = divisor;
		while (Dividend <= Divisor)
		{
			Dividend = Dividend - Divisor;
			result += 1;
		}
	}
    // result = dividend / divisor;
	return result;
    }
};