class Solution {
    public int fib(int n) {
        if (n < 2) return n;
        int first = 0, second = 1;
        int sum = 0;
        for (int i = 1; i < n; i++) {
            sum = first + second;
            first = second;
            second = sum;
        }
        return sum;
    }
}

class Solution {
    public int fib(int n) {
        if (n < 2) {
            return n;
        }
        int[][] q = {{1, 1}, {1, 0}};
        int[][] res = pow(q, n - 1);
        return res[0][0];
    }

    public int[][] pow(int[][] a, int n) {
        int[][] ret = {{1, 0}, {0, 1}};
        while (n > 0) {
            if ((n & 1) == 1) {
                ret = multiply(ret, a);
            }
            n >>= 1;
            a = multiply(a, a);
        }
        return ret;
    }

    public int[][] multiply(int[][] a, int[][] b) {
        int[][] c = new int[2][2];
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];
            }
        }
        return c;
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/fibonacci-number/solution/fei-bo-na-qi-shu-by-leetcode-solution-o4ze/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

class Solution {
    public int fib(int n) {
        double sqrt5 = Math.sqrt(5);
        double fibN = Math.pow((1 + sqrt5) / 2, n) - Math.pow((1 - sqrt5) / 2, n);
        return (int) Math.round(fibN / sqrt5);
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/fibonacci-number/solution/fei-bo-na-qi-shu-by-leetcode-solution-o4ze/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。