LeetCode Divide Two Integers 不使用除号取模乘号实现两数相除

Divide Two Integers

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

不使用乘法，除法和模操作实现除法运算。

思路就是用被除数减去除数，减尽为止，如下面程序。

int divide(int dividend, int divisor) 
	{
		//这里的dividend如果不加long long或者加double， 就会当是INT_MIN的时候溢出
		long long a = abs((long long)dividend);
		long long b = abs((long long)divisor);

		int result = 0;
		while (a >= b)
		{
			a -= b;
			result++;
		}
		return (dividend>>31 ^ divisor>>31)? -result:result;
	}

不过很可惜，上面的程序是无法通过的，因为需要优化一下。上面的程序是基本思想，所以，先要知道这个程序。

下面是两个优化的程序，都可以AC。

int divide2(int dividend, int divisor) 
	{
		//这里的dividend如果不加long long或者加double， 就会当是INT_MIN的时候溢出
		long long a = abs((long long)dividend);
		long long b = abs((long long)divisor);

		int result = 0;
		while (a >= b)
		{
			int count = 0;
			while (a >= b<<(count+1)) count++;
			a -= b<<count;
			result += 1<<count;
		}
		return (dividend>>31 ^ divisor>>31)? -result:result;
	}

	int divide3(int dividend, int divisor)
	{
		long long a = abs((long long)dividend);
		long long b = abs((long long)divisor);

		int result = 0;
		while (a >= b)
		{
			int count = 0;
			while (a >= b<<count) 
			{
				result += 1<<count;
				a -= b<<count;
				count++;
			}			
		}
		return (dividend > 0) ^ (divisor > 0) ? -result : result;
	}

2014-1-25 update

//忘记考虑溢出问题！！！How can I forget???!!!
	int divide(int dividend, int divisor) 
	{
		long long a = dividend;
		long long b = divisor;
		a = abs(a); b = abs(b);
		int res = 0;
		while (a>=b)
		{
			long long t = b;
			for (int i = 1; a >= t; i <<= 1, t <<= 1)
			{
				a -= t;
				res += i;
			}
		}
		return ((dividend<0)^(divisor<0))? -res:res;
	}