My Java development and learning journey ------ & gt; find the string that appears the most frequently and the number of occurrences in the string ------

My Java development and learning journey ------> Find the string that appears the most frequently and the number of times it appears ------

Kingsoft company interview questions: a string may contain ~ Multiple characters in z, such as String data = "aavzcadfdsfsdhshgWasdfasDf ", find the letter and number of occurrences that appear the most frequently. If there are multiple duplicates, find them.

The solution is as follows:

The Code is as follows:

Import java. util. arrayList; import java. util. collections; import java. util. hashMap; import java. util. iterator; import java. util. map; import java. util. treeSet; public class SortTest {public static void main (String [] args) {String input = "httpblogcsdnnetouyangpeng"; new SortTest (). doString (input);} public void doString (String input) {/*** Step 1: * use TreeSet to quickly find all the strings that appear * sort the input strings in ascending order * // convert the String to the character array char [] chars = input. toCharArray (); ArrayList <String> lists = new ArrayList <String> (); // The TreeSet is an ordered set, elements in the TreeSet will be sorted in ascending order // you can quickly find all the strings that appear in the TreeSet <String> set = new TreeSet <String> () through the non-repetition of the TreeSet (); for (int I = 0; I <chars. length; I ++) {lists. add (String. valueOf (chars [I]); set. add (String. valueOf (chars [I]);} // set = [a, B, c, d, e, g, h, l, n, o, p, s, t, u, y] System. out. println ("set =" + set); // sort Collections. sort (lists); // lists = [a, B, c, d, e, e, g, h, l, n, o, o, p, p, s, t, u, y] System. out. println ("lists =" + lists); // convert the sorted character array to StringBufferStringBuffer sb = new StringBuffer (); for (int I = 0; I <lists. size (); I ++) {sb. append (lists. get (I);} input = sb. toString (); // input = abcdeeggghlnnnnooppstttuySystem. out. println ("input =" + input);/*** Step 2: find the occurrence of the same character and record the number of times * // The maximum number of times that a duplicate occurs int max = 0; // The repeated character String maxString = ""; /* // repeated Character List ArrayList <String> maxList = new ArrayList <String> (); * // It is used to save the most frequently occurring strings and times HashMap <String, integer> hm = new HashMap <String, Integer> (); // traverses Iterator <String> its = set. iterator (); while (its. hasNext () {String OS = its. next (); // The first position where the character appears in the sorted input. int begin = input. indexOf (OS); // the last position where the character appears in the sorted input int end = input. lastIndexOf (OS); // number of occurrences of characters int value = end-begin + 1; if (value> = max) {max = value; maxString = OS; hm. put (maxString, max) ;}}for (Map. entry <String, Integer> enties: hm. entrySet () {if (enties. getValue () = max) {System. out. print ("the most repeated letters are:" + enties. getKey (); System. out. println ("the most repeated times are:" + enties. getValue ());}}}}

The running result is as follows:

Set = [a, B, c, d, e, g, h, l, n, o, p, s, t, u, y] lists = [a, B, c, d, e, e, g, h, l, n, o, o, p, p, s, t, t, t, u, y] input = abcdeeggghlnnnnooppstttuy the most repeated letters are: n the most repeated times are: 4

When the number of string duplicates is the same, you can print them all. For example

public static void main(String[] args) {String input = "abbcccddddeeeeeffffffaaaaabbb";new SortTest().doString(input);}

The running result is as follows:

Set = [a, B, c, d, e, f] lists = [a, B, c, d, e, f, f] input = aaaaaabbbbbcccddddeeeeeffffff the most repeated letters are: f the most repeated letters are: 6 the most repeated letters are: a most repeated times are: 6

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Author: Ouyang Peng: Welcome to repost. sharing with others is the source of progress!

Reprinted Please retain the original address: http://blog.csdn.net/ouyang_peng

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