面试题 10- I. 斐波那契数列

F(0) = 0,   F(1) = 1
F(N) = F(N - 1) + F(N - 2), 其中 N > 1.

输入：n = 2
输出：1

输入：n = 5
输出：5

class Solution:
    def fib(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, a + b
        return a % 1000000007

class Solution {
    public int fib(int n) {
        int a = 0, b = 1;
        for (int i = 0; i < n; ++i) {
            int c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return a;
    }
}

class Solution {
public:
    int fib(int n) {
        int a = 0, b = 1;
        for (int i = 0; i < n; ++i) {
            int c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return a;
    }
};

/**
 * @param {number} n
 * @return {number}
 */
var fib = function (n) {
  let a = 0,
    b = 1;
  for (let i = 0; i < n; ++i) {
    const c = (a + b) % (1e9 + 7);
    a = b;
    b = c;
  }
  return a;
};

func fib(n int) int {
    a, b := 0, 1
    for i := 0; i < n; i++ {
        a, b = b, (a + b) % 1000000007
    }
    return a
}