斐波那契數列 爬樓梯問題 python & php版

https://leetcode-cn.com/problems/climbing-stairs/

爬樓梯問題

假設你正在爬樓梯, 需要 n 階你才能到達樓頂
每次你可以爬 1 或 2 個臺階, 你有多少種不同的方法可以爬到樓頂呢?

• 設爬 n 個臺階有 f(n) 種可能
• 假設先爬1階, 剩下 n-1 階有 f(n-1) 種可能
• 假設先爬2階, 剩下 n-2 階有 f(n-2) 種可能
• 因此爬n階可以轉化爲兩種爬n-1階問題之和: f(n) = f(n-1) + f(n-2)

斐波那契公式
$F_{n} = 1 / \sqrt{5} \left [ \left (\frac{1+\sqrt{5}}{2} \right)^{n} - \left (\frac{1-\sqrt{5}}{2} \right)^{n} \right ]$
解方程後得到
$F_{n} = 1 / \sqrt{5} \left ( \left (\frac{1+\sqrt{5}}{2} \right)^{n+1} - \left (\frac{1-\sqrt{5}}{2} \right)^{n+1} \right )$

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python

#動態規劃
def climbStairs(n: int) -> int:
    memory = {}
    memory[1] = 1
    memory[2] = 2
    for i in range(3, n+1):
        memory[i] = memory[i-1] + memory[i-2]
    return memory[i]

#動態規劃 (節省空間)
def climbStairs2(n: int) -> int:
    temp1 = 1
    temp2 = 2
    temp = 0
    for i in range(3, n+1):
        temp = temp1 + temp2
        temp1 = temp2
        temp2 = temp
    return temp

#斐波那契公式
def climbStairs3(n: int) -> int:
    import math
    sqrt5 = 5**0.5
    fibn = math.pow((1+sqrt5)/2, n+1) - math.pow((1-sqrt5)/2, n+1)
    return int(fibn/sqrt5)

step = climbStairs3(10)
print(step)

php

function climbStairs(int $n): int{
    $memory = [];
    $memory[1] = 1;
    $memory[2] = 2;
    for($i=3; $i<$n+1; $i++){
        $memory[$i] = $memory[$i-1] + $memory[$i-2];
    }
    return $memory[$n];
}

function climbStairs2(int $n): int{
    $temp = 0;
    $temp1 = 1;
    $temp2 = 2;
    for($i=3; $i<$n+1; $i++){
        $temp = $temp1 + $temp2;
        $temp1 = $temp2;
        $temp2 = $temp;
    }
    return $temp;
}

function climbStairs3(int $n): int{
    $sqrt5 = sqrt(5);
    $fibn = pow((1+$sqrt5)/2, $n+1) - pow((1-$sqrt5)/2, $n+1);
    return (int)($fibn/$sqrt5);
}

$step = climbStairs3(10);
echo '<pre>';
echo $step;

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