deploy: 2e398c6

Author: acbin

en/lc/2225/index.html

@@ -75871,7 +75871,7 @@ <h3 id="solution-1-hash-table-sorting">Solution 1: Hash Table + Sorting</h3>
 <p>We use a hash table <code>cnt</code> to record the number of matches each player has lost.</p>
 <p>Then we traverse the hash table, put the players who lost 0 matches into <code>ans[0]</code>, and put the players who lost 1 match into <code>ans[1]</code>.</p>
 <p>Finally, we sort <code>ans[0]</code> and <code>ans[1]</code> in ascending order and return the result.</p>
-<p>The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of matches.</p>
+<p>The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the number of matches.</p>
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en/search/search_index.json

@@ -80473,7 +80473,7 @@ <h3 id="_3">方法一：哈希表 + 排序</h3>
 <p>我们用一个哈希表 $\text{cnt}$ 记录每个玩家输掉的比赛场次。</p>
 <p>然后遍历哈希表，将输掉 $0$ 场比赛的玩家放入 $\text{ans}[0]$，将输掉 $1$ 场比赛的玩家放入 $\text{ans}[1]$。</p>
 <p>最后将 $\text{ans}[0]$ 和 $\text{ans}[1]$ 按照升序排序，返回结果。</p>
-<p>时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为比赛场次数。</p>
+<p>时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为比赛场次数。</p>
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