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59 lines
2.1 KiB
HTML
59 lines
2.1 KiB
HTML
<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>
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<p>请实现 <code>KthLargest</code> 类：</p>
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<ul>
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<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>
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<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>
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</ul>
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<p> </p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong><br />
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<span class="example-io">["KthLargest", "add", "add", "add", "add", "add"]<br />
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[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]</span></p>
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<p><strong>输出：</strong><span class="example-io">[null, 4, 5, 5, 8, 8]</span></p>
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<p><strong>解释：</strong></p>
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<p>KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);<br />
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kthLargest.add(3); // 返回 4<br />
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kthLargest.add(5); // 返回 5<br />
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kthLargest.add(10); // 返回 5<br />
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kthLargest.add(9); // 返回 8<br />
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kthLargest.add(4); // 返回 8</p>
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<p> </p>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong><br />
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<span class="example-io">["KthLargest", "add", "add", "add", "add"]<br />
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[[4, [7, 7, 7, 7, 8, 3]], [2], [10], [9], [9]]</span></p>
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<p><span class="example-io"><b>输出：</b>[null, 7, 7, 7, 8]</span></p>
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<p><strong>解释：</strong></p>
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KthLargest kthLargest = new KthLargest(4, [7, 7, 7, 7, 8, 3]);<br />
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kthLargest.add(2); // 返回 7<br />
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kthLargest.add(10); // 返回 7<br />
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kthLargest.add(9); // 返回 7<br />
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kthLargest.add(9); // 返回 8</div>
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<p> </p>
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<strong>提示：</strong>
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<ul>
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<li><code>0 <= nums.length <= 10<sup>4</sup></code></li>
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<li><code>1 <= k <= nums.length + 1</code></li>
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<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
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<li><code>-10<sup>4</sup> <= val <= 10<sup>4</sup></code></li>
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<li>最多调用 <code>add</code> 方法 <code>10<sup>4</sup></code> 次</li>
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</ul>
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