Leetcode 509. Fibonacci Number , Java 解法

最新推荐文章于 2022-08-05 12:55:20 发布

YuanTheCoder

最新推荐文章于 2022-08-05 12:55:20 发布

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分类专栏： Leetcode EASY

本文链接：https://blog.youkuaiyun.com/YuanTheCoder/article/details/95855024

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博客围绕斐波那契数列展开，介绍了递归、迭代、动态规划（自上而下和自下而上）四种解法，并给出了各解法的时间复杂度和空间复杂度，还提供了一篇关于初步理解动态规划的博文链接。

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

Note:

0 ≤ N ≤ 30.

Approach #1 Recursive 我的解法

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

Time complexity: O(2^n)- since T(n) = T(n-1) + T(n-2)is an exponential time
Space complexity: O(n) - space for recursive function call stack

Approach #2 Iterative

class Solution 
{
    public int fib(int N)
    {
        if(N <= 1)
            return N;
        
		int a = 0, b = 1;
		
		while(N-- > 1)
		{
			int sum = a + b;
			a = b;
			b = sum;
		}
        return b;
    }
}

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

Approach #3 Dynamic Programming top-down

class Solution 
{
    int[] fib_cache = new int[31];
	
	public int fib(int N)
    {
        if(N <= 1)
            return N;
        else if(fib_cache[N] != 0)
            return fib_cache[N];
		else 
            return fib_cache[N] = fib(N - 1) + fib(N - 2);
    }
}

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

Approach #4 Dynamic Programming Bottom Up

class Solution 
{
    public int fib(int N)
    {
        if(N <= 1)
            return N;

		int[] fib_cache = new int[N + 1];
		fib_cache[1] = 1;

		for(int i = 2; i <= N; i++)
		{
			fib_cache[i] = fib_cache[i - 1] + fib_cache[i - 2];
		}
		return fib_cache[N];
    }
}

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

另：关于初步理解动态规划很好的一篇博文 https://blog.youkuaiyun.com/u013309870/article/details/75193592