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Basic Javascript (110/111) | Use Recursion to Create a Countdown | freeCodeCamp
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Basic Javascript (110/111) | Use Recursion to Create a Countdown | freeCodeCamp
In a previous challenge, you learned how to use recursion to replace a for loop. Now, let's look at a more complex function that returns an array of consecutive integers starting with 1 through the number passed to the function.
As mentioned in the previous challenge, there will be a base case. The base case tells the recursive function when it no longer needs to call itself. It is a simple case where the return value is already known. There will also be a recursive call which executes the original function with different arguments. If the function is written correctly, eventually the base case will be reached.
For example, say you want to write a recursive function that returns an array containing the numbers 1 through n. This function will need to accept an argument, n, representing the final number. Then it will need to call itself with progressively smaller values of n until it reaches 1.
At first, this seems counterintuitive since the value of n decreases, but the values in the final array are increasing. This happens because the push happens last, after the recursive call has returned. At the point where n is pushed into the array, countup(n - 1) has already been evaluated and returned [1, 2, ..., n - 1].
We have defined a function called countdown with one parameter (n). The function should use recursion to return an array containing the integers n through 1 based on the n parameter. If the function is called with a number less than 1, the function should return an empty array. For example, calling this function with n = 5 should return the array [5, 4, 3, 2, 1]. Your function must use recursion by calling itself and must not use loops of any kind.
In a previous challenge, you learned how to use recursion to replace a for loop. Now, let's look at a more complex function that returns an array of consecutive integers starting with 1 through the number passed to the function.
As mentioned in the previous challenge, there will be a base case. The base case tells the recursive function when it no longer needs to call itself. It is a simple case where the return value is already known. There will also be a recursive call which executes the original function with different arguments. If the function is written correctly, eventually the base case will be reached.
For example, say you want to write a recursive function that returns an array containing the numbers 1 through n. This function will need to accept an argument, n, representing the final number. Then it will need to call itself with progressively smaller values of n until it reaches 1.
At first, this seems counterintuitive since the value of n decreases, but the values in the final array are increasing. This happens because the push happens last, after the recursive call has returned. At the point where n is pushed into the array, countup(n - 1) has already been evaluated and returned [1, 2, ..., n - 1].
We have defined a function called countdown with one parameter (n). The function should use recursion to return an array containing the integers n through 1 based on the n parameter. If the function is called with a number less than 1, the function should return an empty array. For example, calling this function with n = 5 should return the array [5, 4, 3, 2, 1]. Your function must use recursion by calling itself and must not use loops of any kind.
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