1. 递归求N的阶乘

//递归求n的阶乘
public class test {//求n的阶乘方法public static int func(int n) {if (n == 1) {return 1;}int tmp = n * func(n-1);return tmp;}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();System.out.println(func(n));}
}

2. 按顺序打印一个数字的每一位

例如1234打印出1 2 3 4

//按顺序打印一个数字的每一位
public class test {//顺序打印数字public static void func(int num) {if (num > 9) {func(num / 10);}System.out.print(num % 10 + " ");}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int num = sc.nextInt();func(num);}
}

3. 递归求n的和

//递归求n的和
public class test {//public static int  func(int num) {if (num == 1) {return 1;}int sum = num + func(num -1);return sum;}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int num = sc.nextInt();System.out.println(func(num));}
}

4. 计算一个数每一位之和

输入一个非负数，返回组成它的数字之和。例如，输入1729，则应该返回1+7+2+9，它的和是19

//计算一个数每一位之和
public class test {//public static int sum(int num) {if (num < 10) {return num;}return num%10 + sum(num/10);}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int num = sc.nextInt();System.out.println(sum(num));}
}

5. 求斐波那契数列的第N项

//求斐波那契数列的第N项
public class test {//递归方式public static int fib1(int n) {if (n ==1 | n == 2) {return 1;}return fib1(n-1) + fib1(n-2);}//迭代方式（非递归）public static int fib2(int n) {if (n == 1 || n ==2) {return 1;}int f1 = 1;int f2 = 1;int f3 = 0;for (int i = 3; i <= n; i++) {f3 = f1 + f2;f1 = f2;f2 = f3;}return f3;}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();System.out.println(fib1(n));}
}

6. 汉诺塔

//汉诺塔问题
public class test {//打印移动的情况public void move(char pos1, char pos2) {System.out.println(pos1+"->"+pos2+" ");}//hanoiTower，n为当前盘子个数，pos1为起始位置，pos2中转位置，pos3目的位置public static void hanoiTower(int n, char pos1, char pos2, char pos3) {if(n == 1) {move(pos1, pos2);} else {hanoiTower(n-1, pos1,pos3,pos2);move(pos1,pos3);hanoiTower(n-1,pos2,pos1,pos3);}}public static void main(String[] args) {hanoiTower(1,'A','B','C');    // A->BSystem.out.println();hanoiTower(2,'A','B','C');    // A->C A->C B->ASystem.out.println();hanoiTower(3,'A','B','C');    // A->B A->B C->A A->C B->C B->C A->BSystem.out.println();}
}

7. 青蛙跳台阶

斐波那契数列的变形，第一个是1，第二个是2（斐波那契前两个为1）

//青蛙跳台阶问题
public class test {public static int jumpFloor(int target) {if (target == 1) {return 1;}if (target == 2) {return 2;}return jumpFloor(target - 1) + jumpFloor(target - 2);}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();System.out.println(jumpFloor(n));}
}