codewars练习（javascript）

codewars练习（javascript）"/>

codewars练习（javascript）

文章目录

• • codewars-js练习
  • • 2021/2/3
    • • github 地址
      • 【1】<7kyu>【Gauß needs help! (Sums of a lot of numbers).】
      • 【2】<7kyu>【Sum of odd numbers】

codewars-js练习

2021/2/3

github 地址

my github地址，上面有做的习题记录，不断更新…

【1】<7kyu>【Gauß needs help! (Sums of a lot of numbers).】

求和

example：

f(n=100) // returns 5050 
f('n')//false
f(3.14)// false, "n must be an integer"
f(-10)//false, "a positive value of n is required
f(0)// false, "a value of n greater than 0 is required"

solution

<script type="text/javascript">function f(n){console.log(n);if(typeof n == 'number' && n % 1 ==0 && n > 0){var sum =0;for(var i=0;i<=n;i++){sum +=i;}return sum;}return false;	}// 验证console.log(f(100));//5050console.log(f('n'));//fasleconsole.log(f(3.14));//false</script>

【2】<7kyu>【Sum of odd numbers】

Given the triangle of consecutive odd numbers:

             13     57     9    1113    15    17    19
21    23    25    27    29
...

Calculate the row sums of this triangle from the row index (starting at index 1) e.g.:

example：

rowSumOddNumbers(1); // 1
rowSumOddNumbers(2); // 3 + 5 = 8
rowSumOddNumbers(42);//74088

思路

根据n确定第一个数和该行最后一个数，从而计算。

solution

<script type="text/javascript">function rowSumOddNumbers(n) {// console.log(n);var one = n * (n-1) + 1;var last = (n-1) * (n+2) + 1;// console.log(one)var sum = 0;for(var i=one;i<=last;i=i+2){sum +=i;}return sum;}// 验证console.log(rowSumOddNumbers(1)); // 1console.log(rowSumOddNumbers(2)); // 3 + 5 = 8console.log(rowSumOddNumbers(42));//74088</script>

以上为自己思路供大家参考，可能有更优的思路。