【C++】【LeetCode】29. Divide Two Integers

题目

Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.

思路

这题是参考了leetcode上的solution。
这题比较重要的是long long的运用。因为不能使用乘法，除法和取余，所以使用位移。首先排除极端情况，除数为0或者是结果溢出。然后，通过不断位移来找到跟被除数最接近的小于它的除数的2倍数，然后用被除数减去该数来求出差值，循环往复，直到差值小于或者等于除数。

代码

class Solution {public:    int divide(int dividend, int divisor) {        int sign = (dividend<0) ^ (divisor<0) ? -1:1;        if (divisor == 0 || (dividend == INT32_MIN && divisor == -1)) {            return INT32_MAX;        }        long long did = labs(dividend);        long long dis = labs(divisor);        int n = 0;        while (dis <= did) {            int multiple = 1;             long long tmp = dis;            while ((tmp << 1) <= did) {                tmp <<= 1;                multiple <<= 1;            }            did -= tmp;            n += multiple;        }        return sign == 1 ? n:-n;    }};