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。

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。

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;
    }
}