update

leetcode-cn/originData/data-stream-as-disjoint-intervals.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>&nbsp;给你一个由非负整数&nbsp;<code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub></code> 组成的数据流输入，请你将到目前为止看到的数字总结为不相交的区间列表。</p>\n\n<p>实现 <code>SummaryRanges</code> 类：</p>\n\n<div class=\"original__bRMd\">\n<div>\n<ul>\n\t<li><code>SummaryRanges()</code> 使用一个空数据流初始化对象。</li>\n\t<li><code>void addNum(int val)</code> 向数据流中加入整数 <code>val</code> 。</li>\n\t<li><code>int[][] getIntervals()</code> 以不相交区间&nbsp;<code>[start<sub>i</sub>, end<sub>i</sub>]</code> 的列表形式返回对数据流中整数的总结。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例：</strong></p>\n\n<pre>\n<strong>输入：</strong>\n[\"SummaryRanges\", \"addNum\", \"getIntervals\", \"addNum\", \"getIntervals\", \"addNum\", \"getIntervals\", \"addNum\", \"getIntervals\", \"addNum\", \"getIntervals\"]\n[[], [1], [], [3], [], [7], [], [2], [], [6], []]\n<strong>输出：</strong>\n[null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]\n\n<strong>解释：</strong>\nSummaryRanges summaryRanges = new SummaryRanges();\nsummaryRanges.addNum(1);      // arr = [1]\nsummaryRanges.getIntervals(); // 返回 [[1, 1]]\nsummaryRanges.addNum(3);      // arr = [1, 3]\nsummaryRanges.getIntervals(); // 返回 [[1, 1], [3, 3]]\nsummaryRanges.addNum(7);      // arr = [1, 3, 7]\nsummaryRanges.getIntervals(); // 返回 [[1, 1], [3, 3], [7, 7]]\nsummaryRanges.addNum(2);      // arr = [1, 2, 3, 7]\nsummaryRanges.getIntervals(); // 返回 [[1, 3], [7, 7]]\nsummaryRanges.addNum(6);      // arr = [1, 2, 3, 6, 7]\nsummaryRanges.getIntervals(); // 返回 [[1, 3], [6, 7]]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= val &lt;= 10<sup>4</sup></code></li>\n\t<li>最多调用&nbsp;<code>addNum</code> 和 <code>getIntervals</code> 方法 <code>3 * 10<sup>4</sup></code> 次</li>\n</ul>\n</div>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>如果存在大量合并，并且与数据流的大小相比，不相交区间的数量很小，该怎么办?</p>\n",
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