g0501_0600.s0509_fibonacci_number.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0509_fibonacci_number;

// #Easy #Dynamic_Programming #Math #Recursion #Memoization #Dynamic_Programming_I_Day_1
// #Level_1_Day_10_Dynamic_Programming #Udemy_Dynamic_Programming
// #2022_07_25_Time_0_ms_(100.00%)_Space_41.4_MB_(15.60%)

/**
 * 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.
 *
 * **Constraints:**
 *
 * *   `0 <= n <= 30`
**/
public class Solution {
    private int[] memo = new int[31];

    public int fib(int n) {
        if (n == 0) {
            return 0;
        }
        if (n == 1) {
            return 1;
        }
        if (memo[n] != 0) {
            return memo[n];
        }
        memo[n] = fib(n - 1) + fib(n - 2);
        return memo[n];
    }
}