# 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.

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