class Solution {
public:
int divide(int dividend, int divisor) {
if(dividend == 0) return 0;
if(divisor == 1) return dividend;
if(divisor == -1){
if(dividend>INT_MIN) return -dividend;// 只要不是最小的那个整数,都是直接返回相反数就好啦
return INT_MAX;// 是最小的那个,那就返回最大的整数啦
}
long a = dividend;
long b = divisor;
int sign = 1;
if((a>0&&b<0) || (a<0&&b>0)){
sign = -1;
}
a = a>0?a:-a;
b = b>0?b:-b;
long res = div(a,b);
if(sign>0)return res>INT_MAX?INT_MAX:res;
return -res;
}
int div(long a, long b){ // 似乎精髓和难点就在于下面这几句
if(a<b) return 0;
long count = 1;
long tb = b; // 在后面的代码中不更新b
while((tb+tb)<=a){
count = count + count; // 最小解翻倍
tb = tb+tb; // 当前测试的值也翻倍
}
return count + div(a-tb,b);
}
};