coder-xiaomo/leetcode-problemset
1
0
Fork 0
mirror of https://gitee.com/coder-xiaomo/leetcode-problemset synced 2025-09-13 03:11:42 +08:00
Code Issues Projects Releases Wiki Activity GitHub Gitee

批量更新数据

Browse Source
This commit is contained in:
小墨
2025-01-09 20:29:41 +08:00
parent 04ecea043d
commit 48cdd06c2b
5053 changed files with 156164 additions and 135322 deletions
Download Patch File Download Diff File Expand all files Collapse all files

64
leetcode-cn/problem (Chinese)/数据流中的第 K 大元素 [kth-largest-element-in-a-stream].html
View File

@@ -1,40 +1,58 @@
<p>设计一个找到数据流中第 <code>k</code> 大元素的类（class）。注意是排序后的第 <code>k</code> 大元素，不是第 <code>k</code> 个不同的元素。</p>
<p>请实现 <code>KthLargest</code> 类：</p>
<p>请实现 <code>KthLargest</code>&nbsp;类：</p>
<ul>
<li><code>KthLargest(int k, int[] nums)</code> 使用整数 <code>k</code> 和整数流 <code>nums</code> 初始化对象。</li>
<li><code>int add(int val)</code> 将 <code>val</code> 插入数据流 <code>nums</code> 后，返回当前数据流中第 <code>k</code> 大的元素。</li>
</ul>
<p> </p>
<p>&nbsp;</p>
<p><strong>示例：</strong></p>
<p><strong class="example">示例 1：</strong></p>
<pre>
<strong>输入：</strong>
["KthLargest", "add", "add", "add", "add", "add"]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
<strong>输出：</strong>
[null, 4, 5, 5, 8, 8]
<div class="example-block">
<p><strong>输入：</strong><br />
<span class="example-io">["KthLargest", "add", "add", "add", "add", "add"]<br />
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]</span></p>
<strong>解释：</strong>
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3); // return 4
kthLargest.add(5); // return 5
kthLargest.add(10); // return 5
kthLargest.add(9); // return 8
kthLargest.add(4); // return 8
</pre>
<p><strong>输出：</strong><span class="example-io">[null, 4, 5, 5, 8, 8]</span></p>
<p> </p>
<p><strong>解释：</strong></p>
<p>KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);<br />
kthLargest.add(3); // 返回 4<br />
kthLargest.add(5); // 返回 5<br />
kthLargest.add(10); // 返回 5<br />
kthLargest.add(9); // 返回 8<br />
kthLargest.add(4); // 返回 8</p>
<p>&nbsp;</p>
</div>
<p><strong class="example">示例&nbsp;2：</strong></p>
<div class="example-block">
<p><strong>输入：</strong><br />
<span class="example-io">["KthLargest", "add", "add", "add", "add"]<br />
[[4, [7, 7, 7, 7, 8, 3]], [2], [10], [9], [9]]</span></p>
<p><span class="example-io"><b>输出：</b>[null, 7, 7, 7, 8]</span></p>
<p><strong>解释：</strong></p>
KthLargest kthLargest = new KthLargest(4, [7, 7, 7, 7, 8, 3]);<br />
kthLargest.add(2); // 返回 7<br />
kthLargest.add(10); // 返回 7<br />
kthLargest.add(9); // 返回 7<br />
kthLargest.add(9); // 返回 8</div>
<p>&nbsp;</p>
<strong>提示：</strong>
<ul>
<li><code>1 <= k <= 10<sup>4</sup></code></li>
<li><code>0 <= nums.length <= 10<sup>4</sup></code></li>
<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
<li><code>-10<sup>4</sup> <= val <= 10<sup>4</sup></code></li>
<li><code>0 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
<li><code>1 &lt;= k &lt;= nums.length + 1</code></li>
<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
<li><code>-10<sup>4</sup> &lt;= val &lt;= 10<sup>4</sup></code></li>
<li>最多调用 <code>add</code> 方法 <code>10<sup>4</sup></code> 次</li>
<li>题目数据保证，在查找第 <code>k</code> 大元素时，数组中至少有 <code>k</code> 个元素</li>
</ul>