# 509. Fibonacci-Number

difficulty: Easy

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)`.

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**Example 1:**

<pre><code><strong>Input: 2
</strong><strong>Output: 1
</strong><strong>Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: 3
</strong><strong>Output: 2
</strong><strong>Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: 4
</strong><strong>Output: 3
</strong><strong>Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
</strong></code></pre>

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**Note:**

0 ≤ `N` ≤ 30.

## Method One

```Java
class Solution {
    public int fib(int N) {
        //事实上我们只需要三个储存空间就行；
        if( N == 0 ) {
            return 0;
        }
        int prev2 = 0;
        int prev = 1;
        int ans = 1;
        for(int i = 2; i < N + 1; i++ ) {
            ans = prev + prev2;
            prev2 = prev;
            prev = ans;
        }
        return ans;
    }
​
//  以下比较原始，因为空间浪费多；
//     public int fib(int N) {
//         if( N == 0 ) {
//             return 0;
//         }
        
//         int[] ans = new int[N + 1];
//         ans[0] = 0;
//         ans[1] = 1;
//         for(int i = 2; i < N + 1; i++) {
//             ans[i] = ans[i - 1] + ans[i - 2];
//         }
//         return ans[N];
//     }
}
​
```

Rolling array 的方法！ 这比较适用于需要 prevPrev, prev 指针的 dp.

```java
class Solution {
    public int fib(int n) {
        if (n == 0) return 0;
        int[] fibo = new int[]{0, 1, 1};
        for (int i = 3; i <= n; i++) {
            fibo[i % 3] = fibo[(i - 1) % 3] + fibo[(i - 2) % 3];
        }
        return fibo[n % 3];
    }
}
```
