<LeetCode OJ> 371. Sum of Two Integers

Total Accepted: 3722 Total
Submissions: 6898 Difficulty: Easy

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.

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Bit Manipulation

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(M) Add Two Numbers

分析：

说实话每次看到位运算我就头疼。其实不会此问题，纯属学习！虽然知道用位运算，知道与或非的原理！

来自分析区

I have been confused about bit manipulation for a very long time. So I decide to do a summary about it here.

"&" AND operation, for example, 2 (0010) & 7 (0111) => 2 (0010)

"^" XOR operation, for example, 2 (0010) ^ 7 (0111) => 5 (0101)

"~" NOT operation, for example, ~2(0010) => -3 (1101) what??? Don't get frustrated here. It's called two's complement.

1111 is -1, in two's complement

1110 is -2, which is ~2 + 1, ~0010 => 1101, 1101 + 1 = 1110 => 2

1101 is -3, which is ~3 + 1

so if you want to get a negative number, you can simply do ~x + 1

Reference:

https://en.wikipedia.org/wiki/Two%27s_complement

https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html

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)

class Solution {
public:
int getSum(int a, int b) {
if (a == 0) return b;
if (b == 0) return a;
int carry=0,add=0;
while (b != 0) {
add = a ^ b;  //模拟加运算
carry = (a & b) << 1;  //获取进位
a=add;//保存本次计算结果
b=carry;
}

return a;
}
};

可以很容易地用“异或”和“或”操作模拟实现整数加法运算：对应位数的“异或操作”可得到该位的数值，对应位的“与操作”可得到该位产生的高位进位，如：a=010010，b=100111，计算步骤如下：

第一轮：a^b=110101，(a&b)<<1=000100， 由于进位（000100）大于0，

则进入下一轮计算，a=110101，b=000100，a^b=110001，(a&b)<<1=001000，

由于进位大于0，则进入下一轮计算：a=110001，b=001000，a^b=111001，(a&b)<<1=0，

进位为0，终止，计算结果为：111001。

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

原文地址：http://blog.csdn.net/ebowtang/article/details/51812032

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：/article/3664871.html