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.
这道题是CareerCup上的一道原题，难道现在LeetCode的新题都是到处抄来的么，讲解可以参见我之前的博客18.1 Add Two Numbers。简而言之就是用异或算不带进位的和，用与并左移1位来算进位，然后把两者加起来即可，先来看递归的写法如下：
解法一：
class Solution { public: int getSum(int a, int b) { if (b == 0) return a; int sum = a ^ b; int carry = (a & b) << 1; return getSum(sum, carry); } };
上面的解法可以精简到一行，哈哈，叼不叼？
解法二：
class Solution { public: int getSum(int a, int b) { return b == 0 ? a : getSum(a ^ b, (a & b) << 1); } };
也可以写成迭代的样子，思路都是一样的~
解法三：
class Solution { public: int getSum(int a, int b) { while (b) { int carry = (a & b) << 1; a = a ^ b; b = carry; } return a; } };
本文转自博客园Grandyang的博客，原文链接：两数之和[LeetCode] Sum of Two Integers ，如需转载请自行联系原博主。