509. 斐波那契数列（ Fibonacci Number）- 简单

• 题目描述
• 解题方法
• • 解法一：动态规划
  • 解法二：动态规划（优化空间复杂度）
  • 解法三： 递归（暴力解法）

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/fibonacci-number
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
public:
    int fib(int n) {
        if (n <= 1) return n;
        vector 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];
    }
};

• 时间复杂度 O(n)
• 空间复杂度 O(n)

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

• 时间复杂度 O(n)
• 空间复杂度 O(1)

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

• 时间复杂度 O(2^n)
• 空间复杂度 O(n)