371. Sum of Two Integers

题目：

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.

链接：

https://leetcode.com/problems/sum-of-two-integers/#/description

3/14/2017

做错了

抄别人的，并附带解释：

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.理解：carry里面有所有需要进位的bit

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)

把整个解释全读下来，似乎理解了一些。

首先bit运算里or是没有进位的，所以我们将carry单独作为一个数来对待。

因为没有进位，所以xor的时候就是在两个数相加并且没有carry时候的和。（0+0=0， 0+1=1，1+0=0）

所以两数相加也就变成了carry和没有carry的和的相加。

因为carry作用在高一位，所以b = carry << 1

另外因为carry是来自于低位的，所以整个过程中carry的位数逐渐左移，并最后为0（b != 0的判断条件)

所以a就慢慢在循环中吸收了carry，成为最后的和。

**当然，还需要考虑负数相加，等我慢慢来想想。

public class Solution {
    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;
    }
}

介绍bit manipulation非常好的评论：

https://discuss.leetcode.com/topic/49771/java-simple-easy-understand-solution-with-explanation

https://discuss.leetcode.com/topic/50315/a-summary-how-to-use-bit-manipulation-to-solve-problems-easily-and-efficiently/2

以及将来面试前必须要看的网页

https://graphics.stanford.edu/~seander/bithacks.html