JAVA learning lesson 40th (Common Object API)-Map set exercises, apimap

JAVA learning lesson 40th (Common Object API)-Map set exercises, apimap

1. LinkedHashMap set

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> (); // ensure 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 );}}}

Exercise:

Number of occurrences of letters

Import java. util. iterator; import java. util. map; import java. util. treeMap;/* "as6 & dg + dfg-badf453sasxa" record the number of occurrences of letters in the string a (4) B (1) d (4 )... ** train of thought: * It is easy to find that there is a correspondence between letters and numbers, and * The set that can store ing needs to be stored with arrays and maps * to determine whether one side of the relationship is an ordered number? No * Map set * is used to determine whether the uniqueness of a Party has a sequence such as a, B, and c... * Use TreeMap **/public class Main {public static void main (String [] args) {/** 1. converts a string into a character array of a single letter * 2. use each letter as the key to check the table of the Map set * 3. determines whether a letter exists. There is no storage by letter <key>-time <value> (initial time = 1, storage By letter <key>-++ time <value> **/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]); // use the letter as the key to look up the 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. keySet (). iterator (); while (it. hasNext () {Character key = it. next (); Integer value = (Integer) map. get (key); str. append (key + "(" + value + ")");} return str. toString ();}}

Ii. Map set lookup Method

Import java. util. hashMap; import java. util. iterator; import java. util. map; import java. util. imports; import java. util. treeMap; import javax. management. runtimeErrorException; public class Main {public static void main (String [] args) {/* the Map set has many applications in the Lookup Table Method */int num; specified in = new partition (System. in); num = in. nextInt (); String week = Get_Week (num-1); System. out. println (week); System. out. println (Get_Week_English (w Eek); 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 "}; 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"); // pay special attention to the following: map. put ("", Set <Integer>); the Set map can also be placed in the Set. put ("Tuesday", "Tus"); map. put ("Wes", "Wes"); map. put ("Thu"); map. put ("Friday", "Fri"); map. put ("Saturday", "Sta"); map. put ("Sunday", "Sun"); return map. get (week );}}