substring java

same place as the runner1 element, and remember it as walker1. The reason for Walker1 in the front part is that it is impossible to find a larger interval containing the [walker1,runner1] interval so that there is no repetition, because there are already repeating elements in the [Walker1,runner1] interval. So the next target range can only go forward, the lower interval is lost, at this point to the third step;three , the Walker1 front (including itself) lost, the problem back to the first s

1. Find the longest repeating substring of a stringFor example: AAAAAAAAABBBBCCCAAASSSCVVV the longest repeating substring inside this is AAAAAAAAAAlgorithmic thinking: Algorithmic Time complexity (O (n))1. Turn this string into a char array first;2. Iterate over this char array3. Compares the characters of i-1 and I in a char array, if not equal, intercepts the string length, compares it, and replaces it i

each grouping, it is possible to take advantage of the idea of a minimal sliding form, whether the high-speed inference includes the required string.* Visually, it is necessary to start a search from every character, in fact, using two pointers to find the strings that satisfy the condition in S, and each +wordlen. And will not be repeated to the system * meter. Save* A lot of time.Method Two:The idea is still to maintain a form. If the current word is in the dictionary, move on to the right en

.length ());//1. Get the length of a stringChar C1=s1.charat (0);SYSTEM.OUT.PRINTLN (C1);//2. Output kint Index=s1.indexof (' K ');SYSTEM.OUT.PRINTLN (index);//3. Output 0System.out.println (S1.indexof (' z '));//If not present, return-1int Index1=s1.indexof ("be");System.out.println (index1);//4. Gets the position of the first character in the "be" stringSystem.out.println (S1.indexof ("or"));//Return-1int Index2=s1.indexof (' e ', 4);System.out.println (INDEX2)//5. Look back from the specified

++; - } theString subString = s.substring (j+1, k); -ans = substring.length () >ans.length ()?Substring:ans; - } -}algorithmic complexity of the answer : O (N) ~ O (N2)The answer : In two parts, the main method iterates through each character, and then calls the sub-method to expand on both sides by this character, and compares the extended characters of the left and right with two "pointers" .Two sub-methods are called to verify the left

if(Longestlength StartIndex)) { ALongestlength = i-StartIndex; - } - for(intj = StartIndex; J ) { the Hashmap.remove (Char1[j]); - } -StartIndex = lastindex+1; - } + Hashmap.put (Char1[i], i); - } + if(longestlengthStartIndex) Alongestlength=char1.length-StartIndex; at returnlongestlength; - } -}However, after the submission of time Limit exceeded, it seems that the complexity is still too h

Today, we used several methods in the string class to share the code.Title Requirement: Count the number of occurrences of a substring in a specified string (prompting the Java string to appear 6 times)1 Public classsearchsamestring {2 3 Public Static voidMain (string[] args) {4 //define two strings5String shortstr = "java";6String longstr = "Javasd

[0]); system. out. println (A [1]) ;}} Split to separate letters and numbers, simple regular gaps Java code PublicClassTest01 { PublicStaticVoidMain (string [] ARGs ){ String STR = "one123 "; String RegEx = "(? String [] STRs = Str. Split (RegEx ); For(IntI = 0; I System. Out. printf ("STRs [% d] = % S % N", I, STRs [I]); } } } Public class test01 {public static void main (string [] ARGs) {string STR = "one123"; string RegEx = "(?

Index: [Leetcode] leetcode key index (C++/JAVA/PYTHON/SQL)Github:https://github.com/illuz/leetcode030. Substring with concatenation of all Words (hard)links：Title: https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/Code (GitHub): Https://github.com/illuz/leetcodeTest Instructions：Give a string s and a word list, the word length is the same

Topic:Given A stringS, find the longest palindromic substring inS. Assume that maximum length ofSis, and there exists one unique longest palindromic substring.Test instructionsGiven the string s, locate the longest palindrome substring in the string. Assume that the maximum length of this substring is 1000, and that the longest palindrome string is present and un

Package Stringtest;public class StringDemo4 {public static void main (string[] args) {String str1 = "Ilikejavaverymuch"; String str2 = "Java is useful"; StringDemo4 sd4 = new StringDemo4 (); Sd4.sop (Sd4.getmaxsubstring1 (str1, str2));} /** * For the maximum of two strings of the same substring, thought: * 1, the short string, in the way of decreasing the length of the