63 lines
1.8 KiB
Markdown
63 lines
1.8 KiB
Markdown
---
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title: Even Fibonacci Numbers
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localeTitle: 甚至斐波那契数字
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---
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## 问题2：甚至斐波纳契数
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### 方法：
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* 斐波那契序列是其中`fib(n) = fib(n-1) + fib(n-1)`的序列。
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* 在这个挑战中，我们必须将序列中的所有偶数加到`nth`项。
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* `fiboEvenSum(10)` ：
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* 到第10学期的顺序是： 1,2,3,5,8,13,21,34,55,89,144
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* 上述序列中所有偶数的和为： 2 + 8 + 34 + 144 = 188
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### 解：
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#### 基本解决方案 - 迭代：
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```js
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function fiboEvenSum(n) {
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let first = 1, second = 2, sum = 2, fibNum; // declaring and initializing variables
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if (n <= 1) return sum; // edge case
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for (let i = 2; i <= n; i++){ // looping till n
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fibNum = first + second; // getting the ith fibonacci number
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first = second;
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second = fibNum;
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if (fibNum%2 == 0) sum+=fibNum; // If even add to the sum variable
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}
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return sum;
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}
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```
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#### 高级解决方案 - 递归：
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```js
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// We use memoization technique to save ith fibonacci number to the fib array
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function fiboEvenSum(n){
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const fib = [1, 2];
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let sumEven = fib[1];
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function fibonacci(n){
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if (n <= 1) return fib[n]; // base condition
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else if (fib[n]) return fib[n]; // if the number exists in the array we cache it and return
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else {
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fib[n] = fibonacci(n-1) + fibonacci(n-2); // otherwise calculcate and save it to the array
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if (fib[n]%2 == 0) sumEven+=fib[n]; //if the number is even, add it to the sumEven variable
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return fib[n];
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}
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}
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fibonacci(n); // run the recursive function
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return sumEven;
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}
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```
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* [运行代码](https://repl.it/@ezioda004/Project-Euler-Problem-2-Even-Fibonacci-Numbers)
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### 参考文献：
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* [维基百科](https://en.wikipedia.org/wiki/Fibonacci_number) |