59. Divide Two Integers

1. Divide Two Integers My Submissions QuestionEditorial Solution
   Total Accepted: 66073 Total Submissions: 421509 Difficulty: Medium
   Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.
思路：每次减去b(同时b不断的翻倍，从1,2,2^2…)
为什么这样是可行的，因为如果a=nb+x
那么n作为一个整数一定是可以表示成 2m+2m−1+...+21+20
所以从20,21。。。依次减是可行的

#define IMAX numeric_limits<int>::max()
#define IMIN numeric_limits<int>::min()
class Solution {
     public:
          int divide(int dividend,int divisor) {
            assert(divisor!=0);
            int sign = (dividend<0&&divisor>0||dividend>0&&divisor<0)?-1:1;
            unsigned int a = dividend<0?-dividend:dividend;
            unsigned int b = divisor<0?-divisor:divisor;
            unsigned int c = b;
            int multi=0;
            while(a>=b){
                int i=1;    
                while(a>=b){
                    a -=b;
                    multi+=i;
                    if(b<IMAX>>1){
                        b=b<<1;
                        i<<=1;
                    }
                }
                b=c;    
            }
            if(sign>0&&multi==IMIN)return IMAX;
            else return multi*sign;
      }
};