不用+-×÷实现四则运算

我就是我 2022-10-27 13:58 97阅读 0赞

## 不用乘除符号实现加法： ##

思想：用位运算符实现加法运算的过程，注意进位

class Solution { 
    public:
        int add(int a, int b) { 
        	int carry = (unsigned int)(a&b)<<1;    //(进位)
        	int noCarry = a^b;                     //(非进位) 
        	if(carry == 0) return noCarry;         //没有进位，那么直接输出结果
        	else return add(noCarry, carry);       //有进位，则继续加
        }
    };

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDc3MDEyNw_size_16_color_FFFFFF_t_70]

## 不用乘除符号实现乘法： ##

思想：用位运算符实现乘法运算的过程

int multiplication(int a, int b) 
    { 
        int sign = 1;
        if (b < 0){ 
            sign = -sign;
            b = -b;
        }
        if (a < 0){ 
            sign = -sign;
            a = -a;
        }
        int ret = 0;
        
        //核心语句↓↓↓
        while (b){ 
            if (b & 1) 
            { 
                ret = ret + a;
            }
            a <<= 1;
            b >>= 1;
        }
        if (sign < 0) return -ret;
        return ret;
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDc3MDEyNw_size_16_color_FFFFFF_t_70 1]

## 除法 ##

// 可以简单概括为： 60/8 = (60-32)/8 + 4 = (60-32-16)/8 + 2 + 4 = 1 + 2 + 4 = 7
    class Solution { 
    public:
        int divide(int dividend, int divisor) { 
            if(dividend == 0) return 0;
            if(divisor == 1) return dividend;
            if(divisor == -1){ 
                if(dividend>INT_MIN) return -dividend;// 只要不是最小的那个整数，都是直接返回相反数就好啦
                return INT_MAX;// 是最小的那个，那就返回最大的整数啦
            }
            long a = dividend;
            long b = divisor;
            int sign = 1; 
            if((a>0&&b<0) || (a<0&&b>0)){ 
                sign = -1;
            }
            a = a>0?a:-a;
            b = b>0?b:-b;
            long res = div(a,b);
            if(sign>0)return res>INT_MAX?INT_MAX:res;
            return -res;
        }
        int div(long a, long b){   // 似乎精髓和难点就在于下面这几句
            if(a<b) return 0;
            long count = 1;
            long tb = b; // 在后面的代码中不更新b
            while((tb+tb)<=a){ 
                count = count + count; // 最小解翻倍
                tb = tb+tb; // 当前测试的值也翻倍
            }
            return count + div(a-tb,b);
        }
    };

参考1：https://leetcode-cn.com/problems/bu-yong-jia-jian-cheng-chu-zuo-jia-fa-lcof/solution/mian-shi-ti-65-bu-yong-jia-jian-cheng-chu-zuo-ji-7/  
参考2：https://blog.csdn.net/chen134225/article/details/81569009

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