{ "data": { "question": { "questionId": "295", "questionFrontendId": "295", "categoryTitle": "Algorithms", "boundTopicId": 1646, "title": "Find Median from Data Stream", "titleSlug": "find-median-from-data-stream", "content": "
The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value and the median is the mean of the two middle values.
\n\narr = [2,3,4]
, the median is 3
.arr = [2,3]
, the median is (2 + 3) / 2 = 2.5
.Implement the MedianFinder class:
\n\nMedianFinder()
initializes the MedianFinder
object.void addNum(int num)
adds the integer num
from the data stream to the data structure.double findMedian()
returns the median of all elements so far. Answers within 10-5
of the actual answer will be accepted.\n
Example 1:
\n\n\nInput\n["MedianFinder", "addNum", "addNum", "findMedian", "addNum", "findMedian"]\n[[], [1], [2], [], [3], []]\nOutput\n[null, null, null, 1.5, null, 2.0]\n\nExplanation\nMedianFinder medianFinder = new MedianFinder();\nmedianFinder.addNum(1); // arr = [1]\nmedianFinder.addNum(2); // arr = [1, 2]\nmedianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2)\nmedianFinder.addNum(3); // arr[1, 2, 3]\nmedianFinder.findMedian(); // return 2.0\n\n\n
\n
Constraints:
\n\n-105 <= num <= 105
findMedian
.5 * 104
calls will be made to addNum
and findMedian
.\n
Follow up:
\n\n[0, 100]
, how would you optimize your solution?99%
of all integer numbers from the stream are in the range [0, 100]
, how would you optimize your solution?中位数是有序列表中间的数。如果列表长度是偶数,中位数则是中间两个数的平均值。
\n\n例如,
\n\n[2,3,4] 的中位数是 3
\n\n[2,3] 的中位数是 (2 + 3) / 2 = 2.5
\n\n设计一个支持以下两种操作的数据结构:
\n\n示例:
\n\naddNum(1)\naddNum(2)\nfindMedian() -> 1.5\naddNum(3) \nfindMedian() -> 2\n\n
进阶:
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