LeetCode 29:Divide Two Integers

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

If it is overflow, return MAX_INT.
题目要求：利用+,-,移位来实现除法.假设求 dividend / divisor
#include<iostream>
#include<string>
using namespace std;
//最容易的想到的办法，是把除法转化为减法，就像把乘法转化为加法一样，
//提交后发现，这个做法超时了，比如遇到2147483647/3这种时候，
//被除数很大，除数很小，基本都会超时；
//可以利用位运算，每次将除数翻倍，实现加速相减
class Solution{
public:
int divide(int dividend, int divisor){
long long a = abs((long long)dividend); //当dividend=INT_MIN时，-dividend会溢出，所以用long long
long long b = abs((long long)divisor);
long long res = 0;
if (dividend == 0 || divisor == 0) return 0;
if (dividend == INT_MIN && divisor == -1) return INT_MAX;
if (divisor == 1) return dividend;

while (a >= b){
long long c = b;
for (int i = 0; a >= c; i++, c =c << 1){
a -= c;
res += 1 << i;
}
}
return ((dividend^divisor) >> 31) ? -res : res;
//32位int右移31位后剩下的就是符号位，dividend与divisor做异或，判断相除后结果是否为负数
}
};

int main()
{
int a =-2147483648;
int b =2;
Solution sol;
int n = sol.divide(a, b);
cout << n << endl;
system("pause");
return 0;
}