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# A + B Problem

ID: 1; naive

## Solution 1 (Java)

public class Solution {
/**
* @param a: An integer
* @param b: An integer
* @return: The sum of a and b
*/
public int aplusb(int a, int b) {
return a + b;
}
}

## Solution 2 (Java)

public class Solution {
/**
* @param a: An integer
* @param b: An integer
* @return: The sum of a and b
*/
public int aplusb(int a, int b) {
while (b != 0) {
int aTemp = a ^ b;
int bTemp = (a & b) << 1;
a = aTemp;
b = bTemp;
}
return a;
}
}
// 1110 = 14
// 1011 = 11
//11001 = 25
// XOR (不进位加法)
// 1110 00101 10001 a
// 1011 10100 01000 b
// 0101 10001 11001 a ^ b
// AND
// 1110 00101 10001 a
// 1011 10100 01000 b
// 1010 00100 00000 a & b
// a = 1110, 0101, 10001, 11001
// b = 1011, 10100, 01000, 00000

### Notes

In order to avoid using the plus sign in this problem, we can use the XOR operation (if two digits are different, the result is 1; otherwise, the result is 0. The XOR operation is essentially adding without carrying. Thus, we can first XOR the two numbers and then add the carry bits.
This means `a + b = (a ^ b) + (a & b << 1)`
Eventually, the carry bits are zeros (`b == 0`) and we end the operation. Example of tracing the code is shown above. We have 14 + 11 = 25, or in binary: 1110 + 1011 = 11011.