arrays and expanded algorithm

Topic meaning1, given an ordered array A and keyword key, determine whether a key exists, if present, return the subscript value, does not exist then return-1.2, given the unordered array A and keyword key, determine whether a key exists, if there is a return 1, does not exist then return 0.For the 1, 2 problem, we can simply write the O (n) from start to finish scanning algorithm, here is not cumbersome, here we discuss is based on binary search

The following small series will introduce some basic algorithm questions about characters and arrays in JavaScript. I think this is quite good. Now I will share it with you and give you a reference. Let's take a look at the fcc issue recently. Just like upgrading, it attracted me a lot. Today, I took the time to ping the Basic Algorithm Scritping question. Based

This section describes how to implement a key-value table. Linked List Method One way is to use a linked list. The basic idea of this method is to store key-value pairs using a linked list. When you need to find a value, scan the entire linked list. If a matched key exists, the corresponding value is returned. To insert a value, scan the entire linked list. If a matching key is found, the corresponding value is overwritten. If no value is found, add a new key-value pair to the head of the linke

I'm about to graduate.AlgorithmAnd the data structure. Summarize the project-related information. First, review the previous knowledge, and second, record the learning points to facilitate later learning. Description: given an integer array, there are positive and negative numbers in the array. One or more consecutive integers in the array form a subarray. each subarray has an sum, returns the maximum subarray sum of the given array. It also becomes the "maximum sub-segment and" problem. Th

array, and the first pointer points to the last element of the incremented array, while the second pointer points to the first element of the back-face array, and the second pointer points to the smallest element, ending with a loop.The first element mentioned above should be greater than or equal to the last element, special case: If the first 0 elements of the sorted array are moved to the last surface, that is, the array itself, which is still the rotation of the array, so the first number i

This problem is leetcode on the topic, the level of difficulty is 5, just start to do not find a good idea, think it is their IQ is relatively low, and later found really low ...Topic:There is sorted arrays nums1 and nums2 of size m and N respectively.Find The median of the sorted arrays. The overall run time complexity should be O (log (m+n)).Example 1:NUMS1 = [1, 3]nums2 = [2]the median is 2.0Example 2:NU

Algorithm and data structure basics 1: C ++ Implement Dynamic Arrays The anti-spoofing algorithm and data structure are based on the implementation of dynamic arrays. // Array. h # Include # Include # Include Using namespace std; class Array {public: //************************************** * ***

About some basic algorithm questions about characters and arrays in js, and about js array Algorithms Recently, I have been playing the fcc review. It's just like upgrading an instance. It's quite attractive to me. Today, I took the time to ping the Basic Algorithm Scritping question. Based on some tips, it is relatively simple. I would like to learn from the met

Example of how to implement the JS Cartesian Product algorithm and multiple arrays Cartesian Product, js flute This article describes how to implement the Javascript Cartesian Product algorithm and multiple arrays Cartesian product. We will share this with you for your reference. The details are as follows: The impleme

Topic:The array a consists of 10 million random positive integers (int), the design algorithm, given the integer n, finds A and b in a, so that it conforms to the following equation:n = a + bProblem Solving Ideas:1.1000w of random positive integer occupy space is probably 38-40mb, not very large, but still need to consider if the order of magnitude continues to increase the situation. It is best to find a way to not load the array into memory.2. If n

) ^2+ (a.length) ^2+ (some inexplicable comparison algorithm), in short, the complexity of time than the above algorithm is lower. This is the second algorithm, do not put the code.The third algorithm, the idea is to first merge two arrays, and then sort out, repeating the e

Package com.zken.test;/** * @author Iamzken * 2015-8-28 * Merge two ordered arrays * */public class Sorter2 {public static void Merge2sort Edarray (int[] A, int[] B, int[] c) {//a array's current index int i = 0;//b array's current index int j = 0;//c The current index of the array int k = 0;//loop, as long as a and B are not traversed, they always loop whil E (i Copyright NOTICE: This article for Bo Master original article, without Bo Master permiss

(palindrome). Note that you need to remove extra punctuation and whitespace from the string, and then convert the string to lowercase to verify that the string is a palindrome. The value of the function parameter can be"Racecar"，"Racecar"And"Race CAR". function Palindrome (str) {ASTR=str.replace (/[^0-9a-za-z]/g,"'). toLowerCase (); BSTR=astr.split (""). Reverse (). Join (""); if(astr===str) { return true; }Else{ return false; }}palindrome (" Eye");The regular expression can also be:/8Ast

Small algorithm: merge two ordered arrays, which are still sorted after merging. 1/** 2 * merge two ordered arrays, after merging, it is still ordered. 3 * @ Param A: array a 4 * @ Param B: array B 5 * to be merged @ Param C: array C 6 */7 public static void Merge (int A [], int B [], int C []) {8 int lengtha =. length; 9 int lengthb = B. length; 10 11 int Index

(Daily algorithm) LeetCode --- Search in Rotated Sorted Array (Binary Search for rotating arrays)Search in Rotated Sorted Array I II LeetcodePerform binary search for ordered arrays (the following uses a non-decreasing array as an example ): int binarySort(int A[], int lo, int hi, int target) { while(lo { int mid = lo + (hi - lo)/2;

Here we want to compare whether the content in the two arrays is the same. Take the int array as an example. Int [] Arraya = new [] {1, 2, 3, 4, 5} Int [] Arrayb = new [] {5, 3, 2, 1, 4} The values in the preceding two arrays are the same, all of which are 1, 2, 3, 4, 5. The specific algorithm is as follows. The first method is the original method, which uses

Code for sorting arrays by absolute value (c) Address: http://blog.csdn.net/caroline_wendy The sorted array may contain positive and negative values. You can use the semi-query to determine the position of the median, and then use two pointers to sort the two ends in sequence. The solution is: 1. The elements in the array are all positive and return; 2. The elements in the array are all negative and return; 3. If there is a positive number and a neg

,imax = +999999, it will waste a lot of space.Solution Two:With a hash table. Put all the numbers in the array into the table, and for each number, see if their contiguous values ( -1/+1) are also in the table.Code:Import Java.util.hashset;import Java.util.set;public class Test {public static int longestconsecutive (int[] num) {if (num . length = = 0) {return 0;} set18. Spiral MatrixGiven a m*n matrix, the clockwise spiral is printed from the periphery.Cases:[[1, 2, 3],[4, 5, 6],[7, 8, 9]]Output

1 ~ The time complexity of N unordered arrays is O (n) sorting. There are 1, 2 ,.... an unordered Array up to N is used to calculate the sorting algorithm. The time complexity is O (n), the space complexity is O (1), and exchange is used, and only two numbers can be exchanged at a time. (Huawei) Analysis: the array has a certain relationship between values and subscripts, which serves as the terminatio

Find the algorithm for the different data in the two list array!Import java.util.ArrayList;Import Java.util.HashMap;Import java.util.List;Import Java.util.Map;Public class Getlistdifferent {Public static Boolean isdifferentwithlist (String URL, listreturn list.contains (URL)? false:true;} /*** @param list1* @param list2* @return */Public static listLong st = System.nanotime ();maplistFor (String string:list1) {map.put (string, 1); }For (String string: