LintCode1:A+B Problem (位运算经典题)

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本文介绍了一种不使用常规算术运算符实现两数相加的方法，通过位运算完成加法运算，适用于32位整数，包括正数和负数。文章详细解释了算法原理，即a&b代表无进位的和，a^b代表进位，并提供了C++代码实现。

A + B Problem
中文English
Write a function that add two numbers A and B.

Example
Example 1:

Input: a = 1, b = 2
Output: 3
Explanation: return the result of a + b.
Example 2:

Input: a = -1, b = 1
Output: 0
Explanation: return the result of a + b.
Challenge
Of course you can just return a + b to get accepted. But Can you challenge not do it like that?(You should not use + or any arithmetic operators.)

Clarification
Are a and b both 32-bit integers?

Yes.
Can I use bit operation?

Sure you can.
Notice
There is no need to read data from standard input stream. Both parameters are given in function aplusb, you job is to calculate the sum and return it.

解法1：位运算。
a & b代表a和b相加的sum但没有进位。
a ^ b代表a和b相加的sum的进位。
注意：
1）该算法也适用于负数的情形！负数是用补码表示，但该算法也适用。

代码如下：

class Solution {
public:
    /**
     * @param a: An integer
     * @param b: An integer
     * @return: The sum of a and b 
     */
    int aplusb(int a, int b) {
        while (b) {
            int a1 = a ^ b;
            int b1 = (a & b) << 1;
            a = a1;
            b = b1;
        }
        
        return a;    
    }
};