{ "data": { "question": { "questionId": "352", "questionFrontendId": "352", "categoryTitle": "Algorithms", "boundTopicId": 1199, "title": "Data Stream as Disjoint Intervals", "titleSlug": "data-stream-as-disjoint-intervals", "content": "
Given a data stream input of non-negative integers a1, a2, ..., an
, summarize the numbers seen so far as a list of disjoint intervals.
Implement the SummaryRanges
class:
SummaryRanges()
Initializes the object with an empty stream.void addNum(int value)
Adds the integer value
to the stream.int[][] getIntervals()
Returns a summary of the integers in the stream currently as a list of disjoint intervals [starti, endi]
. The answer should be sorted by starti
.\n
Example 1:
\n\n\nInput\n["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]\n[[], [1], [], [3], [], [7], [], [2], [], [6], []]\nOutput\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\nExplanation\nSummaryRanges summaryRanges = new SummaryRanges();\nsummaryRanges.addNum(1); // arr = [1]\nsummaryRanges.getIntervals(); // return [[1, 1]]\nsummaryRanges.addNum(3); // arr = [1, 3]\nsummaryRanges.getIntervals(); // return [[1, 1], [3, 3]]\nsummaryRanges.addNum(7); // arr = [1, 3, 7]\nsummaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]]\nsummaryRanges.addNum(2); // arr = [1, 2, 3, 7]\nsummaryRanges.getIntervals(); // return [[1, 3], [7, 7]]\nsummaryRanges.addNum(6); // arr = [1, 2, 3, 6, 7]\nsummaryRanges.getIntervals(); // return [[1, 3], [6, 7]]\n\n\n
\n
Constraints:
\n\n0 <= value <= 104
3 * 104
calls will be made to addNum
and getIntervals
.102
calls will be made to getIntervals
.\n
Follow up: What if there are lots of merges and the number of disjoint intervals is small compared to the size of the data stream?
\n", "translatedTitle": "将数据流变为多个不相交区间", "translatedContent": " 给你一个由非负整数 a1, a2, ..., an
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。
实现 SummaryRanges
类:
SummaryRanges()
使用一个空数据流初始化对象。void addNum(int val)
向数据流中加入整数 val
。int[][] getIntervals()
以不相交区间 [starti, endi]
的列表形式返回对数据流中整数的总结。\n\n
示例:
\n\n\n输入:\n[\"SummaryRanges\", \"addNum\", \"getIntervals\", \"addNum\", \"getIntervals\", \"addNum\", \"getIntervals\", \"addNum\", \"getIntervals\", \"addNum\", \"getIntervals\"]\n[[], [1], [], [3], [], [7], [], [2], [], [6], []]\n输出:\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解释:\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\n\n
\n\n
提示:
\n\n0 <= val <= 104
addNum
和 getIntervals
方法 3 * 104
次\n\n
进阶: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?
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