Leetcode Divide, integers

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

If It is overflow, return max_int.

Test instructions: Divide by two numbers without multiplication and mod calculations.

Idea: Because each number can be expressed as a binary, that is, num = a*2^0 + b*2^1 ...., so we just have to determine which 2^k can add results, the first way is the loop, the second is to find the maximum 2^k each time, the problem is to change 2 to B just

Class Solution {public:    int divide (int dividend, int divisor) {                long long a = ABS ((double) dividend);               Long Long B = ABS (double) divisor);        Long long res = 0;        while (a >= b) {            long long c = b;            for (int i = 0; a >= C; i++, C <<= 1) {                A-= C;                Res + = 1<<i;            }        }        int flag = ((divisor ^ dividend) >> 31)? -1:1;        if (RES * flag > Int_max)            return int_max;        else return res * flag;}    ;

Class Solution {public:    int divide (int dividend, int divisor) {        long long a = ABS ((double) dividend);               Long Long B = ABS (double) divisor);        Long long res = 0;        while (a >= b) {             long Long T = b, i = 1;             while ((T << 1) < a) {                 T <<= 1;                 I <<= 1;             }             A-= t;             res + = i;        }        int flag = ((divisor ^ dividend) >> 31)? -1:1;        if (RES * flag > Int_max)            return int_max;        else return res * flag;}    ;

Leetcode Divide, integers