62 lines
1.3 KiB
Markdown
62 lines
1.3 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/sum-of-two-integers/description/
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## 题目描述
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```
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Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.
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Example 1:
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Input: a = 1, b = 2
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Output: 3
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Example 2:
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Input: a = -2, b = 3
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Output: 1
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```
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## 思路
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不能使用加减法来求加法。 我们只能朝着位元算的角度来思考了。
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由于`异或`是`相同则位0,不同则位1`,因此我们可以把异或看成是一种不进位的加减法。
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![371.sum-of-two-integers-1](../assets/problems/371.sum-of-two-integers-1.png)
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由于`与`是`全部位1则位1,否则位0`,因此我们可以求与之后左移一位来表示进位。
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![371.sum-of-two-integers-2](../assets/problems/371.sum-of-two-integers-2.png)
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然后我们对上述两个元算结果递归求解即可。 递归的结束条件就是其中一个为0,我们直接返回另一个。
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## 关键点解析
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- 位运算
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- 异或是一种不进位的加减法
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- 求与之后左移一位来可以表示进位
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## 代码
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```js
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/*
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* @lc app=leetcode id=371 lang=javascript
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*
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* [371] Sum of Two Integers
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*/
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/**
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* @param {number} a
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* @param {number} b
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* @return {number}
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*/
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var getSum = function(a, b) {
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if (a === 0) return b;
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if (b === 0) return a;
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return getSum(a ^ b, (a & b) << 1);
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};
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```
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