题干:
中文:
Fibonacci数字,通常表示为F(n),形成一个称为Fibonacci序列的序列,
这样每个数字是前两个数字的总和,从0和1开始。
即, F(0)= 0,F(1)= 1 对于N> 1,F(N)= F(N-1)+ F(N-2)。 给定N,计算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.
英文:
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.
解题思路:
这道题的主要思想是递归。 从N为2开始,每个数是前两个数的和,所以每次返回值为前两个数的和。 当为0和1时是固定值,作为递归的终止条件。 递归一定要有终止条件,要不然可能会导致栈溢出。
package cn.leetcode.easy;
/**
* 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).
*
* @author kimtian
* @date 2019-02-01
* @num 509
*/
public class FibonacciNumber {
/**
* 这道题的主要思想是递归,
* 从N为2开始,每个数是前两个数的和,所以每次返回值为前两个数的和,
* 当为0和1时是固定值,作为递归的终止条件
* 递归一定要有终止条件,要不然可能会导致栈溢出
*
* @param N
* @return
*/
public static int fib(int N) {
//当为0时,结果为0,作为递归的终止条件
if (N == 0) {
return 0;
}
//当为1时,结果为1,作为递归的终止条件
if (N == 1) {
return 1;
}
//其他情况下结果为前两个值之和
return fib(N - 1) + fib(N - 2);
}
/**
* 测试
*/
public static void main(String[] args) {
System.out.println(fib(3));
}
}