【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 ==  || (dividend == INT_MIN && divisor == -)) return INT_MAX;
         long long m = abs((long long)dividend);
         long long n = abs((long long)divisor);  

         int sign = (dividend < ) ^ (divisor < ) ? - : ;
         int res = ;
         while (m >= n) {
             long long tmp = n, p = ;
             while (m >= (tmp << )) {
                 tmp <<= ;
                 p <<= ;
             }
             m -= tmp;
             res += p;
         }
         return sign ==  ? res : -res;
     }
 };

Solution 2

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