Redis3.0 cluster CRC16 algorithm PHP client Implementation method (PHP obtains the Redis partition slot of Redis data in the redis3.0 cluster and obtains the Redis server address of the partition based on the partition slot)

I've been looking at a book like deliberate come, huh? AlgorithmAt that time on Amazon hanging out to pick the book, see the huge number of people recommend this algorithm book, say simple and easy to understand and buy reading. Indeed, the author's ability to describe algorithms is admirable. Let's review the usual algorithms.In turn, I'll take a look at some of the common algorithms that I find interesting, this time it's fast track.Quick Sort Intro

the current problem P scale does not exceed n0, the problem is easy to solve, no longer need to continue decomposition.Adhoc (P) is the basic sub-algorithm in the division method, which can be directly solved for the problem of easy solving.Merge (Y1,y2...yk) is a merging sub-algorithm for solving y1,y2 of P's sub-problem p1,p2,...,pk,..., yk merging to P.The computational efficiency of divide-and-conquer

Association rules are expressions such as a->b, and A and B are the two subkeys that intersect each other in the entire set.The main purpose of mining association rules is to find meaningful correlation relationships in data. Shopping basket analysis is the analysis of customer purchase behavior to discover the relationship between different products.Support degree, confidence level, promotion degreeSupport Degree (a->b) =| ab|/| s|Confidence level (A->B) =| ab|/| a|This

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template A simple example:
#include Run:
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Given an ordered array of arr, adjust arr so that the left half of the array does not have repeating elements and ascending, and does not guarantee that the right is orderedThe partition is OK. The U-Zone is non-repeating and the ascending U is the last position of this region, the initial u=0I do a left-to-right traversal, and on arr[u+1....i] there is no guarantee that there are no duplicate elements and that the ascending area I is the last positio

It is known that a network is required to be divided into subnets of the same size.For example, a known NETWORK:192.168.0.0/23 subnet, which lists all the subnet network addresses according to the subnet block B size to be divided:var network = "192.168.0.0";var CIDR = 23;var B = 4;//Subnet block size here for the change parameter, depending on the size of the partitionvar subnetamount = Math.pow (2, (32-CIDR))/b;//count subnetsvar p = network.split (".");for (i=0;iA = (number (p[0]) + Math.floo

Topic One:Given an ordered array of arr, adjust arr so that the left half of the array does not have a repeating part and ascending, without guaranteeing that the right part is ordered. For example: arr=[1,2,2,2,3,3,4,5,6,6,7,7,8,8,9,9], adjusted after arr=[1,2,3,4,5,6,7,8,9 ...].Requirements:Time complexity O (N), additional space complexity O (1)Program: Public Static void leftunique (int[] arr) {if Null | | Arr.length return;}int u = 0;int i = 1; while (I! = arr.length) {if (arr[i++]! = Arr[u

Private intPartitionint[] nums,intLointHi) { if(lo >=Hi) { returnlo; } inti =lo; intj = Hi + 1; intv =nums[lo]; while(true) { while(nums[++i] V)if(i = = Hi) break; while(nums[--j] >V)if(j = Lo) break; if(i >=J) { break; } intTMP =nums[i]; nums[i]=nums[j]; nums[j]=tmp; } nums[lo]=nums[j]; nums[j]=v; returnj; }Lo and hi are Inclusive. In the returned array, the number of index less than J is equal to the number of J,index greater than J is greater

essentials: Using the vector container,vectorVector, which can be understood as an enhanced version of the array, encapsulates a number of functions that operate on its own.classSolution { Public: intArraypairsum (vectorint>nums) { intRET =0; BOOLFlag =true; Arrayint,20001> hashtable{0 }; for(ConstAuto N:nums) { ++hashtable[n +10000]; } for(inti =0; I 20001;) { if(Hashtable[i] >0) { if(flag) {flag=false; RET+ = (I-10000); --Hashtable[i]; }

sorted sequences, which is used to hold the merged sequence,
set two pointers, The initial position is the starting position of two sorted sequences,
Compare the elements pointed to by two pointers, select a relatively small element into the merge space, And move the pointer to the next position;
Repeat step 3 until a pointer reaches the end of the sequence;
Copies all the remaining elements of another sequence directly to the end of the merge sequence.
a

sorted sequences, which is used to hold the merged sequence,
set two pointers, The initial position is the starting position of two sorted sequences,
Compare the elements pointed to by two pointers, select a relatively small element into the merge space, And move the pointer to the next position;
Repeat step 3 until a pointer reaches the end of the sequence;
Copies all the remaining elements of another sequence directly to the end of the merge sequence.
a

Partition a collection into smaller collectionsThis article describes how to partition a collections into a given number of smaller collections.
Table of Contents
1. Partition a Collection
2. Partition Collection in Java

This is a creation in
Article, where the information may have evolved or changed.
Common internal sorting algorithms are: Insert sort, hill sort, select sort, bubble sort, merge sort, quick sort, heap sort, cardinality sort, etc. Summarize with a picture:
Merge sort (English: merge sort, or mergesort) is an efficient sorting algorithm that creates an O (n log n) on a merge operation. It was first presented by John von Neumann in 1945. The

Topic:Given a linked list and a value x, partition it such that all nodes less than x come before nodes greate R than or equal to x.You should preserve the original relative order of the nodes in each of the.For example,Given 1->4->3->2->5->2 and x = 3,Return 1->2->2->4->3->5 .Idea: Find the two-segment linked list that is divided by x, and connect it together, keeping a watch on the chain header.Code: Public classSolution { PublicListNode

Topic:Given a linked list and a value x, partition it such that all nodes less than x come before nodes greate R than or equal to x.You should preserve the original relative order of the nodes in each of the.For example,Given 1->4->3->2->5->2 and x = 3,Return 1->2->2->4->3->5 .test instructions and analysis: given a linked list and a value x, moving a point in the list with values less than x to a point greater than or equal to X, you need to keep the

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