xqianxin-优快云博客

原创 faiss 编译的坑

编译过程报错：g++ -o demos/demo_ivfpq_indexing -fPIC -m64 -Wall -g -O3 -mavx -msse4 -mpopcnt -fopenmp -Wno-sign-compare -fopenmp demos/demo_ivfpq_indexing.cpp libfaiss.a -g -fPIC -fopenmp /usr/lib64/libope...

2018-02-27 16:37:20 2127

原创 Maximum Subarray 解题

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.For example, given the array [−2,1,−3,4,−1,2,1,−5,4],the contiguous subarray [4,−1,2,1] ha

2016-04-05 22:39:37 252

原创 VS2015 + opencv3.1 环境配置记录

1.安装好 vs2015 和 opencv3.1，假设 opencv3.1 安装在 $(opencv3.1) 中2.配置opencv3.1 的系统环境变量，在path中添加 $(opencv3.1)\build\x64\vc14\bin;3.在 vs2015 的C++项目中，项目——属性——配置属性1）VC++目录——包含目录 : 添加包含目录2）

2016-04-01 12:19:00 585

原创 reverse interger --- leetcode

Reverse digits of an integer.Example1: x = 123, return 321Example2: x = -123, return -321click to show spoilers.Have you thought about this?Here are some good questions to ask before c

2015-04-15 20:48:59 298

原创 Populating Next Right Pointers in Each Node --- leetcode

Given a binary tree struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }Populate each next pointer to point to its next right node.

2015-04-09 23:06:37 246

原创 balanced binary Tree ---leetcode

Given a binary tree, determine if it is height-balanced.For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never diffe

2015-04-09 21:06:15 274

原创一些简单math题

1.Excel Sheet Column Number大意：A->1,B->2,...,AA->27,...解体思路：考虑每位的权值，然后加起来代码：class Solution {public: int titleToNumber(string s) { int n = s.size(); int weidge = 1;

2015-03-23 22:02:53 300

原创 Leetcode刷题日记<Binary Tree Postorder Traversal>

题：Given a binary tree, return the postorder traversal of its nodes' values.解答：一个简单解法，利用栈：1.root入栈2.访问栈顶（加入容器），栈顶出栈，栈顶的左孩子入栈（有的话），栈顶的右孩子入栈*（有的话）3.重复2，直到栈空4.把得到的容器反转即可代码：class Soluti

2014-12-29 22:10:13 222

原创二路归并的一个C++最简单实现

int merge(const int *arr1,int na,const int *arr2,int nb,int* result_arr,int nr){ int i = 0; int j = 0; int k = 0; while (i < na && j < nb) { if (arr1[i]>arr2[j]) { result_arr[k] = arr1[i];

2014-09-15 21:00:40 283