There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. Have you met this question in a real interview?     Example Given A=[1,2,3,4,5,6] and B=[2,3,4,5], the median is 3.5. Given A=[1,2,3] and B=[4…

Given a unsorted array with integers, find the median of it. A median is the middle number of the array after it is sorted. If there are even numbers in the array, return the N/2-th number after sorted. Example Given [4, 5, 1, 2, 3], return 3 Given […

求第k个值 1.归并排序 归并到第k个值为止 时间复杂度:O(k) class Solution { public: // merge-sort to find K-th value double helper(vector<int> A, vector<int> B, int lenA, int lenB, int k) { , j = ; while ((i < lenA) && (j < lenB)) { k--; if (A[i] < B[…

Besides heap, multiset<int> can also be used: class Solution { void removeOnly1(multiset<int> &ms, int v) { auto pr = ms.equal_range(v); ms.erase(pr.first); } void remove(multiset<int> &lmax, multiset<int> &rmin, int v)…

Data Stream Median Numbers keep coming, return the median of numbers at every time a new number added. Have you met this question in a real interview? Example For numbers coming list: [1, 2, 3, 4, 5], return [1, 1, 2, 2, 3]. For numbers coming list:…

Given a unsorted array with integers, find the median of it. A median is the middle number of the array after it is sorted. If there are even numbers in the array, return the N/2-th number after sorted. 思路: 找个sort方法sort完后即可 最简单的就是bubble sort 但是bubble…

Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (2016-02-10) For more problems and solutions, you can see my LintCode repository. I'll keep updating for full summary and better solutions. See cnblogs t…

--------------------------------------------------------------- 本文使用方法:所有题目,只需要把标题输入lintcode就能找到.主要是简单的剖析思路以及不能bug-free的具体细节原因. ---------------------------------------------------------------- ------------------------------------------- 第九周:图和搜索. ---…

Same as LintCode "Sliding Window Median", but requires more care on details - no trailing zeroes. #include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <q…

LintCode 81. Data Stream Median (Hard) 思路: 用一个大根堆保存较小的一半数, 一个小根堆保存较大的一半数. 每次根据num和两个堆顶的数据决定往哪个堆里面放. 放完后进行平衡确保两个堆的size差不超过1. 利用两个堆的size和堆顶值计算median. 大根堆可以表示为priority_queue<int, vector<int>, less<int>>, 其实priority_queue<int>默认就是大根堆.…