Conduct Dot Product of two large Vectors 1. two pointers 2. hashmap 3. 如果没有额外空间,如果一个很大,一个很小,适合scan小的,并且在大的里面做binary search package fb; public class DotProduct { public int dotPro(int[][] v1, int[][] v2) { int[][] shortV; int[][] longV; if (v1.length…

Problem Introduction The dot product of two sequences \(a_1,a_2,\cdots,a_n\) and \(b_1,b_2,\cdots,b_n\) of the same length is equal to \(\sum\limits_{i=1}^na_ib_i=a_1b_1+a_2b_2+\cdots+a_nb_n\) Problem Description Task.The goal is,given two sequences…

These are vectors: They can be multiplied using the "Dot Product" (also see Cross Product). Calculating You can calculate the Dot Product of two vectors this way: a · b = |a| × |b| × cos(θ) Where:|a| is the magnitude (length) of vector a|b| is t…

以下CUDA sample是分别用C++和CUDA实现的点积运算code,CUDA包括普通实现和采用零拷贝内存实现两种,并对其中使用到的CUDA函数进行了解说,code参考了<GPU高性能编程CUDA实战>一书的第十一章,各个文件内容如下: funset.cpp: #include "funset.hpp" #include <random> #include <iostream> #include <vector> #include &…

参考的是<游戏和图形学的3D数学入门教程>,非常不错的书,推荐阅读,老外很喜欢把一个东西解释的很详细. 1.向量点积(Dot Product) 向量点积的结果有什么意义?事实上,向量的点积结果跟两个向量之间的角度有关. 2.向量叉积(Cross Product) 两个向量a,b,它们的叉积表示为axb,这个很容易跟数学中两个数字之间的相乘,但是这里是完全不同的. 两个向量叉积在图形坐标中就很直观了,axb同时垂直与a和b. 我们很容易验证axb是否同时垂直a和b向量.根据向量乘积的知识,我们只…

Given a string, calculate how many substring is palindrome. Ignore non-char characters. Ignore case; Ex: "A@,b123a", equals "Aba", should output 4: "A", "b", "a", "Aba" Spread from center: Like L…

tasks has cooldown time, give an input task id array, output finish time input: AABCA A--ABCA output:7 package fb; import java.util.*; public class Scheduler { public int task(int[] tasks, int cooldown) { int time = 0; HashMap<Integer, Integer> map…

给一组括号,remove最少的括号使得它valid 从左从右各scan一次 package fb; public class removeParen { public static String fix(String str) { StringBuffer res = new StringBuffer(str); int l = 0, r = 0; int i = 0; while (i < res.length()) { if (res.charAt(i) == '(') l++; else…

有个getFriend() API, 让你推荐你的朋友的朋友做你的朋友,当然这个新朋友不能是你原来的老朋友 package fb; import java.util.*; public class ReferFriends { public List<String> recommendation(String name) { List<String> res = new ArrayList<String>(); if (name==null || name.length…

给一个超级大的排好序的vector [abbcccdddeeee]比如,要求返回[{,a}, {,b}, {,c}, {,d}, {,e}......]复杂度要优于O(N) 分析: 如果是binary search找每个char的上下界,worst case要找n次,时间复杂度O(nlogn) 所以考虑每次比较start point和start point + 2^n位置上的数,假如一样就continue,不一样就在区间里面binary search找上界,这样worst case O(N) p…