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;
}
};

参考资料:
http://www.cnblogs.com/grandyang/p/5451942.html
https://leetcode.com/discuss/111682/one-line-java-code
https://leetcode.com/discuss/111601/6-line-java-solution
LeetCode All in One 题目讲解汇总(持续更新中...)
分类: LeetCode
本文介绍了一种不使用加减运算符计算两个整数和的方法,通过异或计算无进位的和,与操作并左移实现进位,结合递归或迭代实现完整算法。
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