# Math / Date

## **sumOfDigit**

**18 :** 1+8 = 9\
**20:** 2+0 = 2



<table><thead><tr><th width="97">方法</th><th>範例</th></tr></thead><tbody><tr><td>笨方法</td><td><pre class="language-javascript"><code class="lang-javascript">let sumOfDigit = 
nums[i].toString()
.split('')
.map(num => Number(num))
.reduce((cur,acc) => cur + acc, 0)
</code></pre></td></tr><tr><td>聰明</td><td><pre class="language-javascript"><code class="lang-javascript">let sum = 0
while (value) {
  sum += value % 10;
  value = Math.floor(value / 10);
  }
</code></pre></td></tr></tbody></table>

笨方法:&#x20;

* 先轉字串 "18"'
* 轉 Array: \["1", "8"]
* 變 Number \[1, 8]
* 再相加 9&#x20;

```javascript
let sumOfDigit = 
nums[i].toString()
.split('')
.map(num => Number(num))
.reduce((cur,acc) => cur + acc, 0)
```

較好方法

```javascript
function getSumofDigit (value){
        let sum = 0
        while (value) {
            sum += value % 10;
            value = Math.floor(value / 10);
        }
        return sum
    }
```

## Math

* \+ , 字串變數字 ''
* Math.pow() 平方，ex. Math.pow(5, 2) = 25
* Math.abs(x) 絕對值, ex Math.abs(-3) = 3&#x20;
* Math.sqrt(x) 平方根 ex. Math.sqrt(9) = 3
* Number.MIN\_SAFE\_INTEGER &#x20;
* Number.MAN\_SAFE\_INTEGER
* Math.random(), ex. 1 - 5 , 1 + Math.floor(Math.random()\*5)

```
Math.floor(Math.random()*4)// 0 - 3
Math.floor(Math.random()*5)// 0 - 4
```

* toString(2) 十進位轉二進位&#x20;

```
let number = 9;
number.toString(2);	// "1001"
```

### Bitwise operators 位元運算子

看起來跟邏輯運算子很像 (&&、 ||)？ 完全不一樣啊！！他是在處理二進位的事，有些題目真的會大大減少撰寫行數．倒是平常寫前端還真的從來沒用到過．

```
// 十轉二
(96).toString(2)
 
// 二轉十
parseInt('1100', 2)
```

*   **& (AND)**

    bit1 & bit2 : 當兩個位元都是1時，回傳1。不是的話就回傳 0&#x20;

```
.    9 (base 10) = 1001 (base 2)
    14 (base 10) = 1110 (base 2)
                   --------------------------------
14 & 9 (base 10) = 1000 (base 2) = 8 (base 10)
```

![](https://1787585077-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LqirD3iAZIDk4oDCfwS%2F-LrPZIi3liVmoab5z-_p%2F-LrPcPTO8c2xbN_5AuQi%2F35%25.jpg?alt=media\&token=e76870c6-6878-49aa-a13c-4d8b4b8d547d)

* **| (OR)**

bit1 | bit2 : 其中一個位元是1時，回傳1。

```
.    9 (base 10) = 1001 (base 2)
    14 (base 10) = 1110 (base 2)
                   --------------------------------
14 | 9 (base 10) = 1111 (base 2) = 15 (base 10)
```

* **^ (XOR)**&#x20;

bit1 ^ bit2 : 當兩個位元一樣時回傳0，不一樣時回傳1。實務運用可以參考 [# 136](https://hannahpun.gitbook.io/leetcode-note/math/136-single-number)

```
     9 (base 10) = 1001 (base 2)
    14 (base 10) = 1110 (base 2)
                --------------------------------
14 ^ 9 (base 10) = 0111 (base 2) = 7 (base 10)
```

* 交换律：a ^ b ^ c  <=> a ^ c ^ b
* 任何数于0异或为任何数 0 ^ n => n
* 相同的数可以等於 0: n ^ n => 0

```
所以  [4,1,2,1,2]
    4 ^ 1 ^ 2 ^ 1 ^ 2
==  1 ^ 1 ^ 2 ^ 2 ^ 4
==  0 ^ 0 ^ 4
==  4
```

## Date



|   |   |   |
| - | - | - |
|   |   |   |
|   |   |   |
|   |   |   |

* new Date()  // Mon Oct 14 2019 15:40:54 GMT-0700 (太平洋夏令時間) {}
* new Date('1990-10-27') // Fri Oct 26 1990 17:00:00 GMT-0700 (太平洋夏令時間) {}

| Method            |                                                                                                                                                                          |
| ----------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |
| getDate()         | 14 幾號                                                                                                                                                                    |
| getDay()          | 1 星期幾                                                                                                                                                                    |
| getFullYear()     | 2019 幾年                                                                                                                                                                  |
| getHours()        | 15 幾點                                                                                                                                                                    |
| getMilliseconds() | 977                                                                                                                                                                      |
| getMinutes()      | 40 幾分                                                                                                                                                                    |
| getMonth()        | 9 幾月                                                                                                                                                                     |
| getSeconds()      | 40 幾秒                                                                                                                                                                    |
| getTime()         | <p>Returns the numeric value of the specified date as the number of milliseconds since January 1, 1970, 00:00:00 UTC (negative for prior times).</p><p>1571093188494</p> |

