題目描述:
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.
網上流傳最廣的方法:
原理:使用異或操作可以進行二進制不帶進位的加減,與操作可以得到進位。
即:
result0 = a ^ b
carry0 = (a & b) << 1
於是有:a + b = result0 + carry0
再進行一次迭代:
result1 = result0 ^ carry0
carry1 = (result0 & carry0) << 1
以此類推:有a + b = result0 + carry0 = result1 + carry1 = result2 + carry2 = ······ = resultN + carryN = result(N+1) ,直到carry(N+1)=0得到結果。
所以步驟是:
1、先讓兩個數字相加,但是不進位,即做異或的操作;
2、計算產生的進位,讓兩個數字位與操作,然後向左移動一位;
3、前兩步的結果相加,重複前兩個步驟直到進位爲0;
很容易寫出代碼:
public class SumOf2Int {
static int getSum(int a, int b) {
int x, y;
while (b != 0){
x = a ^ b;
y = (a & b) << 1;
a = x; b = y;
}
return a;
}
public static void main(String[] args) {
System.out.println(getSum(-15, -5));
}
}