1classSolution {2public:3doublefindkth(vector<int>& nums1,vector<int>& nums2,intk)4{5intm = nums1.size(),n =nums2.size();6if(m >n)7returnfindkth(nums2,nums1,k);//error 1. forget the "return";8if(m ==0)9returndouble(nums2[k -1]);//error 2. write as nums2[n - ...
Unecessary work: Sorted the whole array each time even though we don't need all the non-median numbers to be at its right order in the array. Solution 2. Apply quick select algorithm to select the new median each time a new number is added. Runtime: Sum of O(i) for i: [1, n]...
2dd16af 0001.two-sum 0001.两数之和 0002.两数相加 0003.无重复字符的最长子串 0004.median-of-two-sorted-arrays 0004.寻找两个有序数组的中位数 0005.最长回文子串 0006.Z字形变换 0007.整数反转 0008.字符串转换整数(atoi) 0009.回文数 0011.盛最多水的容器 0012.整数转罗马数字 ...
Can you solve this real interview question? Find Median from Data Stream - 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. * For
double findMedian() - Return the median of all elements so far. For example: addNum(1) addNum(2) findMedian() -> 1.5 addNum(3) findMedian() -> 2 用一个最大堆存放比中位数小(或等于)的元素,用一个最小堆存放比中位数大(或等于)的元素。这里关键的方法是insert(),每当要插入一个元素时...
[leetcode] 295. Find Median from Data Stream @ python 原题Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value. For example, [2,3,4], the median......
0004-median-of-two-sorted-arrays.py 0005-longest-palindromic-substring.py 0007-reverse-integer.py 0010-regular-expression-matching.py 0011-container-with-most-water.py 0012-integer-to-roman.py 0013-roman-to-integer.py 0014-longest-common-prefix.py 0015-3sum.py 0017-lette...
So the median is the mean of the two middle val...***Leetcode 295. Find Median from Data Stream https://leetcode.com/problems/find-median-from-data-stream/description/ 很好的题 要维护小的数都在first里面,需要首先判断要插入的数跟top的对比。 ......
double findMedian()- Return the median of all elements so far. For example: add(1) add(2) findMedian() -> 1.5 add(3) findMedian() -> 2 最大最小堆 复杂度 时间O(NlogN) 空间 O(N) 思路 维护一个最大堆,一个最小堆。最大堆存的是到目前为止较小的那一半数,最小堆存的是到目前为止...
double findMedian() - Return the median of all elements so far. For example: addNum(1) addNum(2) findMedian() -> 1.5 addNum(3) findMedian() -> 2 这里需要我们设计一个数据结构,使得我们可以从当前的字符流中获取中位数。 思路和代码 ...