509. Fibonacci Number #

Problem Statement: #

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.

Constraints: #

• 0 <= n <= 30

Solution: #

public int fib(int n) {
  if(n == 0 || n == 1)
    return n;
  int[] dp = new int[n + 1];
  dp[0] = 0;
  dp[1] = 1;
  for (int i = 2; i < n + 1; i++)
    dp[i] = dp[i - 1] + dp[i - 2];
  return dp[n];
}

public int fib2(int n) {
  if (n <= 1)
    return n;
  return fib(n - 1) + fib(n - 2);
}

... #