LeetCode刷题：509.Fibonacci Number

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.

Note:
0 ≤ N ≤ 30.

解题思路：

• 思路1：采用递归的方法
• 思路2：采用数组，将斐波那契数存入到数组中。

  解答：

  //方法1：
  //效率较低，运行时间20ms
  int fibonacci(int n)
  {
    if(n == 0)
        return 0;
    else if(n == 1)
        return 1;
    return fibonacci(n-1) + fibonacci(n-2); 
  }
  class Solution {
  public:
    int fib(int N) {
        return fibonacci(N);     
    }
  };
  //方法2：
  //运行时间0ms
  class Solution {
  public:
    int fib(int N) {
        int *p = new int[N+1];//开辟大小为N+1的数组
        p[0] = 0;
        p[1] = 1;
        for(int i = 2; i < N+1; ++i)
        {
            p[i] = p[i-1] + p[i-2];
        }
        return p[N];
    }
  };