[LeetCode]74. Divide Two Integers除法运算

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

If it is overflow, return MAX_INT.

 

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解法1：因为题目限制不能用乘法、除法和取模操作，考虑到除法可以使用减法来做，因此可以让被除数不断减去除数，直到小于或者等于0，而循环累计次数就是结果。这样做效率不高，在被除数很小，而除数很大时会超时Time Limit Exceeded

class Solution {
public:
    int divide(int dividend, int divisor) {
        if (divisor == 0) return 0;
        bool isNegative = dividend > 0 ^ divisor > 0;
        long long divd = abs((long long)dividend);
        long long divr = abs((long long)divisor);
        long long loop = 0;
        while (divd > 0) {
            ++loop;
            divd -= divr;
        }
        if (divd < 0) --loop;
        if (loop > INT_MAX) loop = INT_MAX;
        return isNegative ? -loop : loop;
    }
};

注意在调用abs函数时应先将参数提升为long long类型，否则调用int abs(int, int)在输入INT_MIN时会得出错误结果。

 

解法2：考虑使用二分搜索。初始化被减数rem为被除数，减数d为除数，计数值cnt为1，然后减数不断翻倍，计数值自增，直到大于被除数rem，然后被减数减去减数，相当于被除数除了cnt次除数；然后被减数减去这个减数，重设减数和计数值，进行下一次循环，直至被减数小于减数。

class Solution {
public:
    int divide(int dividend, int divisor) {
        if (divisor == 0) return 0;
        if(divisor == 1) return dividend;
        bool isNegative = dividend > 0 ^ divisor > 0;
        long long divd = abs((long long)dividend);
        long long divr = abs((long long)divisor);
        long long loop = 0, rem = divd;
        
        while (rem >= divr) {
            long long d = divr, l = 1;
            while ((d <<= 1) <= rem)
                l <<= 1;
            rem -= (d >> 1);
            loop += l;
        }
        
        if (loop > INT_MAX) loop = INT_MAX;
        return isNegative ? -loop : loop;
    }
};