Huawei computer test algorithm (introductory version)

Summary: Data table records contain table indexes and values (positive integers in the int range). Please merge the records with the same index. That is, sum the values of the same index.

Input requirements: Enter the number of key-value pairs, and then enter pairs of index and value values separated by Spaces

Because the index and value involved in the input must be stored using a map

Output requirements: Output merged key-value pairs (multiple lines)

Input example:

4, 0, 1, 0, 2, 1, 2, 3, 4Copy the code

Example output:

0, 3, 1, 2, 3, 4Copy the code

Code:

import java.util.Map; import java.util.Scanner; import java.util.TreeMap; public class Test1 { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); // To receive inputwhile(scanner.hasNext()) {// Boolean judge int next = scanner.nextint (); // This is the number TreeMap<Integer, Integer> resultMap = new TreeMap(); // Built-in collator, default is key ascending order, meet output requirementsfor (int i = 0; i < next; i++) {
                int key = scanner.nextInt();
                int value = scanner.nextInt();
                if (resultMap.containsKey(key)) {
                    resultMap.put(key, value + resultMap.get(key));
                } else{ resultMap.put(key, value); }}for(Map.Entry entry : Resultmap.entryset ()) {// Iterate through entry=key+ Value entry.getKey() entry.getValue() system.out.println (entry.getKey() +""+ entry.getValue()); }}}}Copy the code

Write a function to count the number of different characters in a string. A character is in the ACSII code range (0 to 127). A newline indicates the end character and is not counted as a character. Those not in the range will not be counted. Multiple identical characters are counted only once. For example, for the string abaca, there are three different characters a, B, and C, so it prints 3.

Input description: Enter a string without Spaces.

Output description: The value contains 0 to 127 characters.

The input

abc
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The output

3
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Just count the number, just like de-weight, and use a set

import java.util.HashSet;
import java.util.Scanner;

public class Test3 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            String input = scanner.next();
            String[] nums = input.split("");
            HashSet countSet = new HashSet();
            for (int i = 0; i < nums.length; i++) {
                if(! countSet.contains(nums[i])) { countSet.add(nums[i]); } } System.out.println(countSet.size()); }}}Copy the code

Anything that involves de-counting can be done with bitmaps. Because each different piece of data only needs to be stored in binary bits, the storage space used for statistics is greatly reduced

class Test3_BitSet {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while(scanner.hasNext()) { String input = scanner.next(); char[] nums = input.toCharArray(); // Bitsets can only hold ints and char, not strings, so input.split()""); Into the input. ToCharArray (); BitSet countSet = new BitSet(128);for (int i = 0; i < nums.length; i++) {
                if(! countSet.get(nums[i])) { countSet.set(nums[i]); } } System.out.println(countSet.cardinality()); // BitSet can only be usedsetGet cardinality, size()}}}Copy the code

Put an English statement in reverse order of words. For example, “I am a boy” is displayed as “boy a am I” in reverse order. All words are separated by a space and contain no other characters except English letters.

Input description: Enter an English statement, each word separated by a space. Ensure that the input contains only Spaces and letters.

Output description: Get the sentence in reverse order.

The input

I am a boy
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The output

boy a am I
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Code (both pieces of code are fine) :

import java.util.Scanner;

//public class Test5 {
//    public static void main(String[] args) {
//        Scanner scanner=new Scanner(System.in);
//        while(scanner.hasNextline ()){// Instead of looping one character at a time, loop one action at a time // String input = scanner.nextline (); [] nums=input.split()""); // StringBuffer StringBuffer =new StringBuffer(); //for(int i=nums.length-1; i>=0; i--){ // stringBuffer.append(nums[i]); //if(i! // stringbuffer.append ();""); // } // } // System.out.print(stringBuffer); // class Test5_Help { Public static void main(String[] args) {Scanner Scanner =new Scanner(system.in);while(scanner.hasNextline ()){// Instead of looping one character at a time, loop one behavior at a time String input = scanner.nextline (); NextLine String[] nums=input.split()""); // Separate with Spacesfor(int i=nums.length-1; i>=0; i--){ System.out.print(nums[i]+""); }}}}Copy the code

All are the entry level, Java write to choose a good set framework, set with its own to remove heavy properties, map with its own key value pair, array with its own alphabetic method, bitSet to reduce space complexity, brush a few master, and then go to see the binary tree, dynamic planning these, ten thousand hours of law should be deliberately learned, rather than simple repetition.

Source click here

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