numpy array 过滤后的数组,索引值从 0 开始. pandas Series 过滤后的 Series ,保持原来的索引,原来索引是几,就是几. 什么意思呢,来看个栗子: import numpy as np import pandas as pd # 有两个相同的数组,一个是pd Series 一个是 np array a = pd.Series([1, 2, 3, 4]) c = np.array([1, 2, 3, 4]) # 通过索引数组来过滤数组 d = a[a>3] e =
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
