Description:
Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.
Example:
Given a = 1 and b = 2, return 3.
思路
不使用算术运算求和那么只能考虑直接在二进制位上进行位运算,事实上利用异或运算(^)和与运算(&)就能完成加法运算要做的事情.
异或运算完成相加但是不进位,而与运算计算出哪些地方需要进位,在通过左移运算(<<)就可以完成进位操作了。
1.循环的方法
class Solution {
public:
int getSum(int a, int b) {
int sum = a;
while (b != 0)
{
sum = a ^ b;
//calculate sum of a and b without thinking the carry
b = (a & b) << 1;
//calculate the carry
a = sum;
//add sum(without carry) and carry
}
return sum;
}
};
2.递归的方法
class Solution {
public:
int getSum(int a, int b) {
if(a && b)
return getSum(a^b, (a&b) << 1);
else
return a|b;
}
};