【LeetCode】029. Divide Two Integers

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

If it is overflow, return MAX_INT.

题解：

　　思路就是被除数减去除数，减尽为止。优化的方法是尽量少的做减法。由于不能用乘法，可以利用位操作，左移一位即为该数乘上2

Solution 1

class Solution {
public:
    int divide(int dividend, int divisor) {
        if (divisor == 0 || (dividend == INT_MIN && divisor == -1)) return INT_MAX;
        long long m = abs((long long)dividend);  
        long long n = abs((long long)divisor);  
        
        int sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1;
        int res = 0;
        while (m >= n) {
            long long tmp = n, p = 1;
            while (m >= (tmp << 1)) {
                tmp <<= 1;
                p <<= 1;
            }
            m -= tmp;
            res += p;
        }
        return sign == 1 ? res : -res;
    }
};

 Solution 2 

class Solution {
public:
    int divide(int dividend, int divisor) {
        if (!divisor || dividend == INT_MIN && divisor == -1) 
            return INT_MAX;
        int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
        unsigned dvd = abs(dividend);
        unsigned dvs = abs(divisor);
        int res = 0;
        while (dvd >= dvs) {
            unsigned temp = dvs, multiple = 1;
            // 不可写作 dvd >= (tmp << 1),因为有可能溢出
            while (dvd - temp >= temp) { 
                temp <<= 1;
                multiple <<= 1;
            }
            dvd -= temp;
            res += multiple;
        }
        return sign * res;
    }
};