斐波那契数列——递归和数组双解决

斐波那契数列:

第n项等于n-1和n-2项之和

Fn=  Fn-1+ Fn-2;

解题:关键代码如下

递归解决:

	public static void main(String[] args) {
		int m = f(18);
		System.out.println(m);
	}
	
	public static int f(int n) {
		if (n == 1 || n == 2) {
			return 1;
		}
		else {
			return f(n - 1) + f(n - 2);
		}
	}

数组解决:(效率高,速度快,占空间小)

	public static void main(String[] args) {
		int[] f = new int[100];
		f[0] = 1;
		f[1] = 1;
		for (int i = 2; i < f.length; i++) {
			f[i] = f[i-1] + f[i-2];
		}
		System.out.println(f[17]);
	}

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转载自blog.csdn.net/likunkun__/article/details/80723831