# 509. Fibonacci Number

The **Fibonacci numbers**, commonly denoted `F(n)` form a sequence, called the **Fibonacci sequence**, such that each number is the sum of the two preceding ones, starting from `0` and `1`. That is,

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

Given `N`, calculate `F(N)`.

**Example 1:**

```
Input: 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
```

**Example 2:**

```
Input: 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
```

**Example 3:**

```
Input: 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
```

#### 解题要点：

创建一个新的数组串，先声明0和1位上的数分别为，0跟1。然后使用dp的方法，轮流加进数组后续数位上的数字，返回index为N的数。

```java
class Solution {
    public int fib(int N) {
        if(N == 0) return 0;
        int[] f = new int[N + 1];
        f[0] = 0;
        f[1] = 1;
        for(int i = 2; i <= N; i++){
            f[i] = f[i-1] + f[i-2];
        }
        return f[N];
    }
}
```

#### 方法二：

根据**Fibonacci**定理，当前数等于前两位数相加，先声明两个初始值0和1，用for loop循环加入到当前数，for loop结束后返回它。

```java
class Solution {
    public int fib(int N) {
        if(N == 0) return 0;
        int first = 0;
        int second = 1;
        int thrid = 1;
        for(int i = 2; i <= N; i++){
            thrid = first + second;
            first = second;
            second = thrid;
        }
        return thrid;
    }
}
```
