<div id="readability-page-1" class="page"><div>
            <h2 id="%EB%AC%B8%EC%A0%9C">문제</h2><figure><a href="https://leetcode.com/problems/array-partition-i/"><div><p>Loading...</p><p>Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.</p><p><img src="https://leetcode.com/favicon-192x192.png"><span>LeetCode</span></p></div><p><img src="https://leetcode.com/static/images/LeetCode_Sharing.png"></p></a></figure><h2 id="%EB%8B%B5%EC%95%88">답안</h2><pre><code>class Solution:
    def arrayPairSum(self, nums: List[int]) -&gt; int:
        sorted_nums = sorted(nums)

        return sum([num for idx, num in enumerate(sorted_nums) if not idx % 2])
</code></pre><p>좀더 간결하게</p><pre><code>class Solution:
    def arrayPairSum(self, nums: List[int]) -&gt; int:
        return sum(sorted(nums)[::2])
</code></pre>
        </div></div>

일시불 프로그래밍

문제
답안
class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        sorted_nums = sorted(nums)
        return sum([num for idx, num in enumerate(sorted_nums) if not idx % 2])
좀더 간결하게
class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        return sum(sorted(nums)[::2])

Leetcode - 561. Array Partition I

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