371. Sum of Two Integers

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.

Credits:
Special thanks to @fujiaozhu for adding this problem and creating all test cases.

这道题让我们实现两数相加，但是不能用加号或者其他什么数学运算符号，那么我们只能回归计算机运算的本质，位操作Bit Manipulation，我们在做加法运算的时候，每位相加之后可能会有进位Carry产生，然后在下一位计算时需要加上进位一起运算，那么我们能不能将两部分拆开呢，我们来看一个例子759+674

1. 如果我们不考虑进位，可以得到323

2. 如果我们只考虑进位，可以得到1110

3. 我们把上面两个数字假期323+1110=1433就是最终结果了

然后我们进一步分析，如果得到上面的第一第二种情况，我们在二进制下来看，不考虑进位的加，0+0=0， 0+1=1, 1+0=1， 1+1=0，这就是异或的运算规则，如果只考虑进位的加0+0=0, 0+1=0, 1+0=0, 1+1=1，而这其实这就是与的运算，而第三步在将两者相加时，我们再递归调用这个算法，终止条件是当进位为0时，我们直接返回第一步的结果，参见代码如下：

For this, problem, for example, we have a = 1, b = 3,

In bit representation, a = 0001, b = 0011,

First, we can use "and"("&") operation between a and b to find a carry.

carry = a & b, then carry = 0001

Second, we can use "xor" ("^") operation between a and b to find the different bit, and assign it to a,

Then, we shift carry one position left and assign it to b, b = 0010.

Iterate until there is no carry (or b == 0)

// Iterative
public int getSum(int a, int b) {
	if (a == 0) return b;
	if (b == 0) return a;

	while (b != 0) {
		int carry = a & b;
		a = a ^ b;
		b = carry << 1;
	}
	
	return a;
}

// Iterative
public int getSubtract(int a, int b) {
	while (b != 0) {
		int borrow = (~a) & b;
		a = a ^ b;
		b = borrow << 1;
	}
	
	return a;
}

// Recursive
public int getSum(int a, int b) {
	return (b == 0) ? a : getSum(a ^ b, (a & b) << 1);
}

// Recursive
public int getSubtract(int a, int b) {
	return (b == 0) ? a : getSubtract(a ^ b, (~a & b) << 1);
}

// Get negative number
public int negate(int x) {
	return ~x + 1;
}