计算a，b之和。但不能使用 + 或者 -
考点，bit运算符
//www.greatytc.com/p/7e00102bf8ad
http://blog.gainlo.co/index.php/2016/07/19/3sum/
”或“运算 |
int a = 2 //10;
int b = 3 //11;
int c = a|b;
System.out.println(c); //result is : 3 /11
“与”运算 &
int a = 2; //10
int b = 3; //11
int c = a&b;
System.out.println(c); //result is : 2 //10
“异或”运算a^b
int a = 2; //10
int b = 3; //11
int c = a^b;
System.out.println(c); //result is : 1 //01
def getSum(a, b):
return a if b == 0 else getSum(a ^ b, (a & b) << 1)
int add(int a , int b)
{
int sum = a;
/*直到进位的结果为0*/
while(b != 0)
{
sum = a ^ b; /*不考虑进位的和，用异或运算求和；*/
b = (a & b) << 1; /*只考虑进位的产生值*/
a = sum;
}//while
return sum;
}
正常101011 - 100001
101011 ^ 100001 = 001010
101011 & 100001 =100001
101011
- 100001
= 001010（不考虑进位）
(101011 & 100001) = 100001（表明有两个位置有进位）
左移一位表示进位。000010，然后一直加直到完全没有进位。
不用额外变量交换两个整数的值
a = 101
b = 110
a = a^b = 011
b = a^b = 101
a = a^b = 110
