leetcode - Divide Two Integers

Divide Two Integers

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

If it is overflow, return MAX_INT.

分析：

1、设法在不溢出的前提下，把除数和被除数转化成正数进行计算。

2、当 dividend/divisor>2 时，利用右移一位实现除数倍增，快速求得商。否则商为1.

class Solution {
public:
    int divide(int dividend, int divisor) {
       if (divisor == 0)
		 throw exception();
	 if (dividend == 0x80000000 && divisor == -1)
		 return 0x7fffffff;
	 if (divisor == 0x80000000)
		 return dividend == 0x80000000 ? 1 : 0;
	 int res = 0;
	 bool isfushu = false;
	 if ((dividend>0 && divisor<0) || (dividend<0 && divisor>0))
		 isfushu = true;
	 if (dividend == 0x80000000)
	 {
		 dividend += abs(divisor);
		 ++res;
	 }
	 dividend = abs(dividend);
	 divisor = abs(divisor);
	 res+=divide_core(dividend,divisor);
	 return isfushu ? -res : res;
    }
    
    int divide_core(int dividend, int divisor)
    {
	 int res = 0;
	 if (divisor == dividend)
		 ++res;
	 else if (divisor<dividend)
	 {
		 int r = divisor, count = 0,intmark=0x80000000;
		 r = r << 1;
		 while ((r&intmark)==0 && r <= dividend)
		 {
			 ++count;
			 r = (r << 1);
		 }
		 if (count>0)
		 {
			 res += (1 << count);
			 dividend -= (divisor << count);
			 res+=divide_core(dividend,divisor);
		 }
		 else
		 {
		     ++res;
		 }
	 }
	 return res;
    }
};