Title:
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
思路就是每次以两倍增长除数,知道即将大于被除数,然后用被除数减去,在循环。
遇到int整数问题一定要考虑到INT_MAX和INT_MIN。
class Solution { public: int divide(int dividend, int divisor) { int final = 0; /////// overflow/////// if (divisor == 0 || (dividend == INT_MIN && divisor == -1)){ return INT_MAX; } int nega = 0; if ((dividend>0&&divisor<0) || (dividend<0&&divisor>0)) nega = 1; //如果为INT_MIN则求abs为越界 long long a = abs((long long)dividend); long long b = abs((long long)divisor); if (b > a) return 0; long long sum = 0;//后面有sum += sum防止越界 int count = 0; while (a >= b) { count = 1; //a >= b保证了最少有一个count sum = b; while (sum + sum <= a){ sum += sum; count += count; } a -= sum; final += count; } if (nega) final = 0 - final; return final; } };