LeetCode 509. Fibonacci Number [Easy]

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Fibonacci Number - LeetCode

题目

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,

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F(0) = 0,   F(1) = 1
F(N) = F(N - 1) + F(N - 2), for N > 1.

Given N, calculate F(N).

Example 1:

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

Example 2:

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

Example 3:

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

Note:

0 ≤ N ≤ 30.

解法1[Java]：递归

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class Solution {
    public int fib(int N) {
        if (N == 0) {
            return 0;
        } else if (N == 1) {
            return 1;
        } else {
            return fib(N - 1) + fib(N - 2);
        }
    }
}

时空复杂度

时间复杂度：

空间复杂度：

解法2[Java]：

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class Solution {
    public int fib(int N) {
        if (N == 0) {
            return 0;
        } else if (N == 1) {
            return 1;
        } else {
            int last = 1;
            int secondToLast = 0;
            int result = 0;
            for (int i = 2; i <= N; ++i) {
                result = last + secondToLast;
                secondToLast = last;
                last = result;
            }
            return result;
        }
    }
}

时空复杂度

时间复杂度：$O(n)$

空间复杂度：$O(1)$

解法3[Java]：

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参考资料

  1. Fibonacci number - Wikipedia
  2. Program for Fibonacci numbers - GeeksforGeeks