题目：
The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,
F(0) = 0, F(1) = 1
F(N) = F(N - 1) + F(N - 2), for N > 1.
Given N, calculate F(N).
Example 1:
Input: 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
Example 2:
Input: 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
Example 3:
Input: 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
Note:
0 ≤ N ≤ 30.
采用递归的方法，秒解这道题：代码中也有详细的描述
public class FibonacciSequence {
//1.函数是用来干什么的：计算f(N)
//2.函数结束条件是什么：当n为0或者为1的时候就应该结束
//3.寻找函数的等式或者给下一层返回的是什么：f(n) = f(n-1) + f(n-2)
public int fib(int N) {
if(N == 0 || N ==1){return N;}
return fib(N-1)+fib(N-2);
}
public static void main(String[] args) {
FibonacciSequence sequence = new FibonacciSequence();
int i = sequence.fib(3);
System.out.println(i);
}
}
做递归题就按照我这种套路来进行做就比较好理解，不要在意递归里面的具体细节，只需关注这个等式，以及停止的一个条件。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/fibonacci-number
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