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