Day16 递归

1、说到递归就不得不说菲波那契数列，F(1)=1，F(2)=1, F(n)=F(n-1)+F(n-2)（n>=3，n∈N*），相当于每次都是自己调用了自己；

2、sumToN() 这个方法是求0-N之和，可以理解成 N 与 sumToN(N-1) 之和，即 sumToN(0) = 0 ， sumToN(N) = sumToN(N - 1) + N（n∈N*）；

 1 public class Recursion {
 2     /**
 3      * Sum to N. No loop, however a stack is used.
 4      * 
 5      * @param paraN The given value.
 6      * @return The sum.
 7      */
 8     public static int sumToN(int paraN) {
 9         if (paraN <= 0) {
10             // Basis.
11             return 0;
12         } // Of if
13 
14         return sumToN(paraN - 1) + paraN;
15     } // Of sumToN
16 
17     /**
18      * Fibonacci sequence.
19      * 
20      * @param paraN The given value.
21      * @return The sum.
22      */
23     public static int fibonacci(int paraN) {
24         if (paraN <= 0) {
25             return 0;
26         } if (paraN == 1) {
27             return 1;
28         }// Of if
29         return fibonacci(paraN - 1) + fibonacci(paraN - 2);
30     }// Of fibonacci
31 
32     public static void main(String args[]){
33         int tempValue = 5;
34         System.out.println("0 sum to " + tempValue + sumToN(tempValue));
35         tempValue = -1;
36         System.out.println("0 sum to " + tempValue + sumToN(tempValue));
37 
38         for (int i = 0; i < 10; i++) {
39             System.out.println("Fibonacci " + i + ": " + fibonacci(i));
40         } // Of for i
41     } // Of main
42 } // Of class Recursion

标签：递归,No,int,paraN,sumToN,Day16,public
来源： https://www.cnblogs.com/f1uency/p/16219608.html