509 - Fibonacci Number

#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).

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

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

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

1. 確定dp[i] 含意， dp[i] => 第i個fib數值
2. 遞推公式 => dp[i] = dp[i-1] + dp[i-2]
3. dp數值如何初始化 => dp[0] = 0 dp[1] = 1
4. 遍歷順序 => 從前向右
5. 打印dp數組

class Solution {
public:
    int recursive( int n ) {
        if ( n == 0 ) return 0;
        if ( n == 1 ) return 1;
        else return recursive( n - 1 ) + recursive( n - 2 );

    }

    int fib(int n) {
        // return recursive( n );  
        if ( n <= 1 ) return n;
        vector<int> dp( n + 1 );
        dp[0] = 0;
        dp[1] = 1;
        for ( int i = 2; i <= n; i++ ) {
            dp[i] = dp[i-1] + dp[i-2];
        }

        return dp[n];
    }
};