leetcode 第4题 中位数技巧: 对于长度为L的有序数组,它的中位数是(a[ceil((L+1)/2)]+a[floor((L+1)/2)])/2 算法原理: 类似三分法求极值 两个人都前进,谁前进之后比较小,就让谁前进. import math class Solution(object): def findpos(self, x, nums1, nums2): a, b = 0, 0 l = len(nums1) + len(nums2) while a + b < x: # prin…
解题关键:模板与思路.面试题 #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> #include<vector> using namespace std; typedef long long ll; //两种方式求on中位数 int partition(int…
求三个数组的中位数,以及中位数的中位数. import java.util.Arrays; public class median { public static void main(String[] args) { //m=3,n=3 long[] a = {1,2,6,4,5,9}; long[] b = {3,9,23,51,5}; long[] c = {13,234,1,54,32}; Arrays.sort(a); //用来排序的方法 Arrays.sort(b); Arrays…
题目难度:hard There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Example 2: n…
一.Description FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of the cows give as much or more than the median; half give as much or less. Given an odd number of cows N (1 <= N < 10…
Median Sum You are given N integers A1, A2, ..., AN. Consider the sums of all non-empty subsequences of A. There are 2N−1 such sums, an odd number. Let the list of these sums in non-decreasing order be S1, S2, ..., S2N−1. Find the median of this list…
#include <iostream> using namespace std; /*函数作用:取待排序序列中low.mid.high三个位置上数据,选取他们中间的那个数据作为枢轴*/ int median(int arr[], int b[], int len1, int low, int high) { int mid = low + ((high - low) >> 1); //计算数组中间的元素的下标 int &lowData = low >= len1 ?…
1029 Median(25 分) Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequences…
实例1: SET @ID = 0; SELECT AVG(loan_amount) from ( SELECT @ID:=@ID+1 as ID, loan_amount FROM table_xxx ORDER BY loan_amount ) a where IF( ROUND(@ID/2,0)%2=0,ID in (ROUND(@ID/2,0),ROUND(@ID/2,0)/2 + 1), ID=ROUND(@ID/2,0) ) 实例2: SET @ID = 0; SELECT AVG…
#include <iostream> #include <cassert> #include <stack> #include <math.h> using namespace std; int QuickSortOnce(int a[], int low, int high) { // 将首元素作为枢轴. int pivot = a[low]; int i = low, j = high; while (i < j) { // 从右到左,寻找首个小…
