29. Divide Two Integers

欢迎fork and star：Nowcoder-Repository-github

29. Divide Two Integers

题目

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

If it is overflow, return MAX_INT. 

解析

• 此题主要考虑整数的处理，主要考虑正负号，是否溢出
• 另外针对int最小负值取绝对值的用法

class Solution_29 { //此题保证一定除尽的条件
public:
	int divide(int dividend, int divisor) {

		if (divisor == 0||(dividend==INT_MIN&&divisor==-1)) //考虑越界的问题
		{
			return INT_MAX;
		}
		int ret = 0;
		bool sign = (dividend > 0) ^ (divisor > 0); //异号为1

		//long long dividend_ = labs(dividend); //转换 abs(-2147483648)=-2147483648 ;换成labs在leetcode AC过但是VS2013编译器还是负数
		//long long divisor_ = labs(divisor);  //区别lab，abs,fabs()

		//long long dividend_ = llabs(long long(dividend)); //转换 abs(-2147483648)=-2147483648 //强制类型转换一个long和int 一样的，必须有两个long long

                long long dividend_ = llabs((dividend)); //用llabs()即可
		long long divisor_ = llabs(divisor);  //区别lab，abs,fabs()
		while (dividend_>=divisor_)
		{
			int temp = divisor_;
			int multi = 1;//被除数的次数
			while (dividend_ >= (temp << 1)) //成倍的增加，时间更快
			{
				temp=temp << 1;
				multi=multi << 1;
			}
			dividend_ -= temp;
			ret += multi;
		}
		
		return sign?-ret:ret;
	}
};

	Solution_29 su_29;
	su_29.divide( 0-2147483648, 1);

题目来源

• 29. Divide Two Integers