剑指offer--面试题9：斐波那契数列--Java实现

题目描述：

写一个函数，输入n，求斐波那契数列的第n项。斐波那契数列定义如下

F(0)=0

F(1)=1

F(n) = F(n-1)+ F(n-2)（n≧2）

思路：

1.递归，但是数值太大容易栈溢出，大量重复计算，效率低下

2.非递归，效率高

import java.math.BigInteger;

public class Fibonacci {

//递归算法
public  static BigInteger fibonacci(int n){
if(n == 1){
return BigInteger.valueOf(1);
}
if(n <= 0){
return BigInteger.valueOf(0);
}

return fibonacci(n - 1).add(fibonacci(n - 2));

}
//非递归算法
public static BigInteger fibonacciAdvance(int n){
int[] result = {0, 1};
if(n < 2){
return BigInteger.valueOf(result
);
}

BigInteger fibNMinusOne = BigInteger.ONE;
BigInteger fibNMinusTwo = BigInteger.ZERO;
BigInteger fibN = BigInteger.ZERO;

for (int i = 2; i <= n; i++) {
//f[i] = f[i-1] + f[i-2]
fibN = fibNMinusOne .add(fibNMinusTwo);
//f[i-2] = f[i-1]
fibNMinusTwo = fibNMinusOne;
//f[i-1] = f[i]
fibNMinusOne = fibN;
}
return fibN;

}
public static void main(String[] args) {

System.out.println(fibonacci(20));
System.out.println(fibonacciAdvance(20));

}

}