[Math]Divide Two Integers

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

If it is overflow, return MAX_INT.

1.除法转换为加法，加法转换为乘法，y个x相加的结果就是x*y.

假设结果为dividend的一半，用此值与divisor相乘，如果相乘结果大于dividend则，high=dividend,然后继续二分

/**
两个整数相除也可能会溢出，最小负数除以-1就溢出了
*/
class Solution {
public:
    long long int multiply(long long x,long long int y)
    {
        if(y==1){
            return x;
        }
        long long int res = multiply(x,y>>1);
        res = (1&y) ? (res<<1) + x: (res<<1);
        return res;
    }
    
    int divide(int dividend, int divisor) {
        long long int l_dividend = fabs(dividend);
        long long int l_divisor = fabs(divisor);    
        
        if(l_dividend < l_divisor){
            return 0;
        }
        if((dividend==INT_MIN && divisor==-1) || divisor==0){
            return INT_MAX;
        }
        
        int sign =  ((dividend>>31)^(divisor>>31)) ? -1:1;
    
        long long int low = 1,high =l_dividend;
       
        while(low<=high){
            long long int  mid = low+((high-low)>>1);
            long long int res = multiply(l_divisor,mid);
            if(res == l_dividend){
                return mid*sign;
            }else if(res<l_dividend){
                low = mid+1;
            }else{
                high = mid-1;
            }
        }
        
        return (low-1)*sign;
    }
};

 

2.除法分配率：a/b = (x+y)/b = x/b+y/b ，其中a=x+y;

42/6;

dividend = 42，divisor=6;

42 = 6*2*2 + 18  ---> 42/6 = 24/6 + 18/6

18 = 6*2 + 6;     ----> 18/6 = 12/6 + 6

6 = 6+0;            ---->  6/6  = 6/6 + 0

所以：42/6 = 2*2+2+1=7；

 

/**
两个整数相除也可能会溢出，最小负数除以-1就溢出了
*/
class Solution {
public:
    int divide(int dividend, int divisor) {
        long long int l_dividend = fabs(dividend);
        long long int l_divisor = fabs(divisor);    
        
        if(l_dividend < l_divisor){
            return 0;
        }
        if((dividend==INT_MIN && divisor==-1) || divisor==0){
            return INT_MAX;
        }
        
        int sign =  ((dividend>>31)^(divisor>>31)) ? -1:1;
        int res = 0;
        while(l_dividend >= l_divisor){
            long long int tmp = l_divisor;
            int occur_times = 0;
            while(tmp <= l_dividend){
                tmp = tmp<<1;
                occur_times++;
            }
            res += (1<<(occur_times-1));
            l_dividend -= (tmp>>1);
        }

        return res*sign;
    }
};

Next challenges: (E) Excel Sheet Column Title (M) Number of Digit One (M) Perfect Squares