# 509.Fibonacci Number

**509.Fibonacci Number**

难度:Easy

> 斐波那契数，通常用 F(n) 表示，形成的序列称为斐波那契数列。该数列由 0 和 1 开始，后面的每一项数字都是前面两项数字的和。也就是：

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

```
示例 1：

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

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

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

提示：
0 ≤ N ≤ 30
```

对于计算过的直接查表即可。

```
class Solution {
public:
    int fib(int N) {
        if(fb.count(N)) return fb[N];
        int res=fib(N-1)+fib(N-2);
        fb[N]=res;
        return res;
    }
private:
    unordered_map<int,int> fb = {\{0,0},{1,1}};
};
```

> 执行用时 : 12 ms, 在Fibonacci Number的C++提交中击败了28.60% 的用户 内存消耗 : 9.1 MB, 在Fibonacci Number的C++提交中击败了100.00% 的用户


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