本文概要
- Java的递归例1：无限次
- Java递归实施例2：有限次
- Java的递归例3：阶乘数
- Java的递归例4：斐波纳契数列
它使代码紧凑，但复杂的理解。
句法：
returntype methodname(){
//code to be executed
methodname();
//calling same method
}Java的递归例1：无限次
public class RecursionExample1 {
static void p(){
System.out.println("hello");
p();
}public static void main(String[] args) {
p();
}
}输出：
hello
hello
...
java.lang.StackOverflowErrorJava递归实施例2：有限次
public class RecursionExample2 {
static int count=0;
static void p(){
count++;
if(count<
=5){
System.out.println("hello "+count);
p();
}
}
public static void main(String[] args) {
p();
}
}输出：
hello 1
hello 2
hello 3
hello 4
hello 5Java的递归例3：阶乘数
public class RecursionExample3 {
static int factorial(int n){
if (n == 1)
return 1;
else
return(n * factorial(n-1));
}public static void main(String[] args) {
System.out.println("Factorial of 5 is: "+factorial(5));
}
}输出：
Factorial of 5 is: 120上述程序的工作：
factorial(5)
factorial(4)
factorial(3)
factorial(2)
factorial(1)
return 1
return 2*1 = 2
return 3*2 = 6
return 4*6 = 24
return 5*24 = 120Java的递归例4：斐波纳契数列
public class RecursionExample4 {
static int n1=0，n2=1，n3=0;
static void printFibo(int count){
if(count>0){
n3 = n1 + n2;
n1 = n2;
n2 = n3;
System.out.print(" "+n3);
printFibo(count-1);
}
}public static void main(String[] args) {
int count=15;
System.out.print(n1+" "+n2);
//printing 0 and 1
printFibo(count-2);
//n-2 because 2 numbers are already printed
}
}【Java递归】输出：
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377
