算法-java实现

1. 质因数分解

public static List<Integer> factorize(int n){
        List<Integer>  factors = new ArrayList();
        int i;
        boolean isPrime = true;
        for (i = 2; i <= Math.sqrt(n); i++){
            if (n % i == 0){
                factors.add(i);
                isPrime = false;
                break;
            }
        }
        if (isPrime){
            factors.add(n);
        }else{
            factors.addAll(factorize( n / i));
        }
        return factors;
    }

 2.题目：求s = a + aa + aaa + aaaa + aa...a的值，其中a是一个数字。例如2 + 22 + 222 + 2222 + 22222(此时共有5个数相加)，

解法1：非递归解法

public static int f2(int a, int n){
        int s = 0;
        int an = 0;
        for (int i = 1; i <= n; i++){
            an += a * Math.pow(10, n-1);
            s += an;
        }
        return s;
    }

解法2：递归解法

 // g = aa...a
    public static int g(int a, int n){
        if (n > 1){
            return g(a, n -1) + a * (int)Math.pow(10, n-1);
        }else if ( n == 1){
            return a;
        }else {
            throw new RuntimeException();
        }
    }
    //f = a + aa + aaa + aaaa + aa...a
    public static int f(int a, int n){
        if (n > 1){
            return f(a, n -1) + g(a,n);
        }else if ( n == 1){
            return a;
        }else {
            throw new RuntimeException();
        }
    }