递归打印出斐波那契前100个

/**
     * 打印出斐波那契前100个
     * 1 1 2 3 5 8 13 21...
     * @param n
     * @return
     */
    public static int f(int n) {
        if (1 == n || 2 == n) {
            return 1;
        }
        return f(n - 2) + f(n - 1);
    }

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

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转载自blog.csdn.net/qq_33101675/article/details/79978247