Java Learning Lesson 40th (Common Object API)-MAP collection Exercises

First, Linkedhashmap Collection

Import Java.util.hashmap;import java.util.iterator;import Java.util.linkedhashmap;import Java.util.Map;public class Main {public static void main (string[] args) {hashmap<integer, string> HM = new linkedhashmap<integer,string> ();//Guaranteed ordered Hm.put (1, "a"); Hm.put (5, "C"); Hm.put (3, "E"); Hm.put (4, "s");iterator<map.entry<integer,string> > It = Hm.entryset (). iterator (); while (It.hasnext ()) {Map.entry<integer, string> meentry = It.next (); Integer Key = Meentry.getkey (); String value = Meentry.getvalue (); System.out.println ("Key-value:" +key+ "-" +value);}}}

Practice:

Record the number of letters appearing

Import java.util.iterator;import java.util.map;import java.util.treemap;/* "As6&dg+dfg-badf453sasxa" Number of occurrences of letters in the record string A (4) B (1) d (4) ... * Ideas: * It is easy to see that there is a correspondence between the letters and the number of times, and that you need to store a set of mappings that can store a mapping. Is the array and map * Determine whether the party in the relationship is an ordinal number? No * need to use map set * to judge the uniqueness of the party with the order as a,b,c ... * So use TreeMap * */public class Main {public static void main (string[] args) {/* * 1. Converts a string into a single-letter character array * 2. Use each letter key to check the table of the map collection. * 3. Determine if the letter exists, there is no letter <key>-time<value> storage (initial time = 1) * exists, by letter <key>-++time<value> The way to store * */string str = "ASDGDFGBADFSASXA"; String result = Get (str); SYSTEM.OUT.PRINTLN (result);} public static string Get (String str) {char[] ch = str.tochararray (); map<character,integer> map = new treemap<character,integer> (); for (int i = 0; i < ch.length; i++) {if (! (') A ' <=ch[i]&&ch[i]<= ' Z ' | | ' A ' <=ch[i]&&ch[i]<= ' Z ') continue;integer value = Map.get (Ch[i]); Look up the letter as a key table int time = 1;if (value!=null) time = value + 1;map.put (Ch[i], time);} Return map_to_string (MAP);} private static String map_to_string (map map) {StringBuilder str = new StringBuilder ();iterator<character> it = MAP.K Eyset (). iterator (); while (It.hasnext ()) {Character key = It.next (); Integer value = (integer) map.get (key); Str.append ( Key+ "(" +value+ ")");} return str.tostring ();}}

Second, map set Check table method

Import Java.util.hashmap;import java.util.iterator;import Java.util.map;import Java.util.scanner;import Java.util.treemap;import Javax.management.runtimeerrorexception;public class Main {public static void Main (string[] args) {/* The application of Map collection in table-checking method is more */int num; Scanner in = new Scanner (system.in); num = In.nextint (); String week = Get_week (num-1); SYSTEM.OUT.PRINTLN (week); System.out.println (Get_week_english (Week)); In.close ();} public static String get_week (int num) {if (num>7 | | num <0) throw new RuntimeException ("input error"); String[] Week = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"};return Week[num];} public static string Get_week_english (String Week) {map<string, string> Map = new hashmap<string, string> (); if (! ( Week.equals ("Monday") | | Week.equals ("Tuesday") | | Week.equals ("Wednesday") | | Week.equals ("Thursday") | | Week.equals ("Friday") | | Week.equals ("Saturday") | | Week.equals ("Sunday")) {throw new RuntimeException ("input wrong Week"); Map.put ("Monday", "Monday");//special note, Map.put ("",set<integer>); set Map.put ("Tuesday", "Tus") can also be placed in the collection; map. put ("Wednesday", "Wes"), Map.put ("Thursday", "Thu"), Map.put ("Friday", "Fri"), Map.put ("Saturday", "Sta"); Map.put ("Sunday", "Sun"); return Map.get (week);}}

Java Learning Lesson 40th (Common Object API)-MAP collection Exercises