509. 斐波那契数

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

输入：2
输出：1
解释：F(2) = F(1) + F(0) = 1 + 0 = 1.

输入：3
输出：2
解释：F(3) = F(2) + F(1) = 1 + 1 = 2.

输入：4
输出：3
解释：F(4) = F(3) + F(2) = 2 + 1 = 3.

func fib(N int) int {
	if N == 0 {
		return 0
	}

	if N <= 2 {
		return 1
	}

	n1, n2 := 1, 1 // n1为n-1，n2为n-2

	for i := 3; i < N; i++ {
		n1, n2 = n1+n2, n1
	}

	return n1 + n2

}

func fib2(n int) int {
       if n==0 ||n==1{
        return n
    }
    dp:=make([]int,n+1)
    dp[0]=0
    dp[1]=1
    for i:=2;i<=n;i++{
        dp[i]=(dp[i-1]+dp[i-2])%1000000007
    }
    return dp[n]
}