A problem about Recursive return value
function fibonacciCatch(num) {
let cache = [1, 1];
(function fibonacci(n) {
if(typeof cache[n] == "number") {
return cache[n]
} else {
// return
return cache[n] = fibonacci(n - 1) + fibonacci(n - 2)
}
})(num - 1);
return cache;
}
console.log(fibonacciCatch(4)) //[ 1, 1, 2, 3 ]
if you do not write the return in the above comments, then you will generate [1Magi 1dagan]
. As far as I understand it, that sentence is equivalent to copying the nth item of the array. After the assignment, it can be used for operation again, so why do you need to return it?
recently, I have been learning algorithms for beginners. Sometimes I always feel like I can"t get around it on the recursive side. Please help me explain it in more detail. Thank you
.
fibonacci (n-1) + fibonacci (n-2) to do this addition fibonacci must return a value, or else it is undefined+undefined
what the landlord should say is that there is return on the inner floor
.
function fibonacciCatch(num) {
let cache = [1, 1];
(function fibonacci(n) {
if(typeof cache[n] == 'number') {
return cache[n]
} else {
cache[n] = fibonacci(n - 1) + fibonacci(n - 2)
}
})(num - 1);
return cache;
}
if the function fibonacci does not return a value, then execute
fibonacci(n - 1) + fibonacci(n - 2)
In this sentence, the two functions are added several times, so
NaN is returned.
fibonacciCatch(6)
[1, 1, 2, NaN, NaN, NaN]
The value in the
cache array (starting with the third item) is the return value from the function fibonacciCatch. If you do not have the return function, the default return value is
of undefined.
if you do not add return , the valid code when running FobonacciCatch (4) is equivalent to:
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