Java Challenge - factorial

Given n of 1 or more, return the factorial of n, which is n * (n-1) * (n-2) ... 1. Compute the result recursively (without loops). factorial(1) → 1 factorial(2) → 2 factorial(3) → 6

Hisoka:

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public int factorial(int n) { if(n == 1) return 1; return n*factorial(n-1); }

Hasil:

Codingbat:

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public int factorial(int n) { // Base case: if n is 1, we can return the answer directly if (n == 1) return 1; // Recursion: otherwise make a recursive call with n-1 // (towards the base case), i.e. call factorial(n-1). // Assume the recursive call works correctly, and fix up // what it returns to make our result. return n * factorial(n-1); }

Woooh...sama yaaak... :D

Java Challenge - stringX - warm_up_2

Given a string, return a version where all the "x" have been removed. Except an "x" at the very start or end should not be removed. stringX("xxHxix") → "xHix" stringX("abxxxcd") → "abcd" stringX("xabxxxcdx") → "xabcdx"

Hisoka:

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public String stringX(String str) { if(str.length() > 1 && str.charAt(0) == 'x' && str.charAt(str.length()-1) == 'x') { return "x"+str.replace("x","")+"x"; }else if(str.contains("x") && str.length()!=1) { return str.replace("x", ""); }else { return str; } }

Hasil:

Codingbat:

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public String stringX(String str) { String result = ""; for (int i=0; i<str.length(); i++) { // Only append the char if it is not the "x" case if (!(i > 0 && i < (str.length()-1) && str.substring(i, i+1).equals("x"))) { result = result + str.substring(i, i+1); // Could use str.charAt(i) here } } return result; }