[LeetCode April Challange] Day 15 - 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: 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

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Solution

Recursion

Time complexity : O(2^n)
Space complexity : O(n)

class Solution {
public:
    int fib(int n) {
        if (n < 2) return n;
        return fib(n-2) + fib(n-1);
    }
};

Dynamic Programming

Time complexity : O(n)
Space complexity : O(n)

class Solution {
public:
    int fib(int n) {
        vector<int> dp(n+2, 0);
        dp[0] = 0;
        dp[1] = 1;
        for (int i=2; i<=n; ++i)
            dp[i] = dp[i-2]+dp[i-1];
        return dp[n];
    }
};

Iterative

Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    int fib(int n) {
        if (n < 2) return n;
        int prev = 0;
        int curr = 1;
        while (1<n--) {
            int next = prev + curr;
            prev = curr;
            curr = next;
        }
        return curr;
    }
};