mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-11-04 19:53:12 +08:00
42 lines
1.3 KiB
HTML
42 lines
1.3 KiB
HTML
<p>Given an integer <code>k</code>, <em>return the minimum number of Fibonacci numbers whose sum is equal to </em><code>k</code>. The same Fibonacci number can be used multiple times.</p>
|
|
|
|
<p>The Fibonacci numbers are defined as:</p>
|
|
|
|
<ul>
|
|
<li><code>F<sub>1</sub> = 1</code></li>
|
|
<li><code>F<sub>2</sub> = 1</code></li>
|
|
<li><code>F<sub>n</sub> = F<sub>n-1</sub> + F<sub>n-2</sub></code> for <code>n > 2.</code></li>
|
|
</ul>
|
|
It is guaranteed that for the given constraints we can always find such Fibonacci numbers that sum up to <code>k</code>.
|
|
<p> </p>
|
|
<p><strong class="example">Example 1:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> k = 7
|
|
<strong>Output:</strong> 2
|
|
<strong>Explanation:</strong> The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ...
|
|
For k = 7 we can use 2 + 5 = 7.</pre>
|
|
|
|
<p><strong class="example">Example 2:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> k = 10
|
|
<strong>Output:</strong> 2
|
|
<strong>Explanation:</strong> For k = 10 we can use 2 + 8 = 10.
|
|
</pre>
|
|
|
|
<p><strong class="example">Example 3:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> k = 19
|
|
<strong>Output:</strong> 3
|
|
<strong>Explanation:</strong> For k = 19 we can use 1 + 5 + 13 = 19.
|
|
</pre>
|
|
|
|
<p> </p>
|
|
<p><strong>Constraints:</strong></p>
|
|
|
|
<ul>
|
|
<li><code>1 <= k <= 10<sup>9</sup></code></li>
|
|
</ul>
|