[LeetCode] Divide Two Integers

[Problem]
Divide two integers without using multiplication, division and modoperator.

[Analysis]
注意以下几点：
（1）被除数和除数为0
（2）被除数或除数为0
（3）正负号，确定正负号后，最好将被除数和除数都转换为正数操作
（4）将int dividend 和 int divisor转换成unsigned，避免溢出
（5）不能使用乘法、除法，可以使用移位运算
（6）还有一些小trick，看代码就懂了，你懂的。

[Solution]

class Solution {
public:
   
    // 2^n
    int power(int n){
        int table[17] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536};
   
        if(n <= 17 && n >= 0){
            return table[n];
        }
        else{
            return -1;
        }
    }
   
    // divide
    int divide(int dividend, int divisor) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
       
        // both of them are 0
        if(dividend == 0 && divisor == 0){
            return 1;
        }
        // dividend is 0
        else if(dividend == 0){
            return 0;
        }
        else if(divisor == 0){
            // invalid
            return -1;
        }
   
        // set mark
        int mark = 1;
        if((dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0)){
            mark = -1;
        }
   
        // reset dividend and divisor
        if(dividend < 0){
            dividend = 0 - dividend;
        }
        if(divisor < 0){
            divisor = 0 - divisor;
        }
       
        // be careful overflow
        unsigned int a = dividend;
        unsigned int b = divisor;
        int res = 0;
        while(a >= b){
            if(b < 1000000){ // a trick
                for(int i = 10; i >= 0; --i){
                    int tmp = b << i;
                    if(a >= tmp){
                        while(a >= tmp){
                            res += power(i);
                            a -= tmp;
                        }
                        break;
                    }
                }
            }
            else{
                res++;
                a -= b;
            }
        }
       
        // negtive
        if(mark == -1){
            return 0 - res;
        }
        // positive
        else{
            return res;
        }
    }
};

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