华为笔试
公众号介绍：Microstrong(小强)同学主要研究机器学习、深度学习、计算机视觉、智能对话系统相关内容，分享在学习过程中的读书笔记！期待您的关注，欢迎一起学习交流进步！
知乎专栏：https://zhuanlan.zhihu.com/Microstrong
Github：https://github.com/Microstrong0305
个人博客：https://blog.csdn.net/program_developer
1. 给定一行字符串，求出这行字符串中出现频率最高的字符，字符串中含有标点符号，字符不区分大小写。如果出现频率相同时，输出先出现在字符串中的字符。
输入：
输入一行字符串，字符串中可能包含多个空格，也可能包含标点符号，但肯定包含字符。
输出：
输出字符的大写和出现频率。
输入样例：
Abcdefg ahigkl Mnopq rstu o v wBBBBBB!
输出样例：
B7
已经AC代码：
import java.util.LinkedHashMap;
import java.util.Map.Entry;
import java.util.scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
String strs = input.replaceAll(" ","").toLowerCase();
LinkedHashMap<Character, integer> map = new LinkedHashMap<Character, Integer>();
Character CharStr = null;
Integer CountmaxLength = 0;
for (Character temp : strs.toCharArray()) {
if (map.containskey(temp)) {
map.put(temp, map.get(temp) + 1);
} else {
map.put(temp, 1);
}
}
for (Entry<Character, Integer> entry : map.entrySet()) {
if (entry.getValue() > CountmaxLength) {
CharStr = entry.getKey();
CountmaxLength = entry.getValue();
}
}
System.out.println(Character.toUpperCase(CharStr)+CountmaxLength.toString());
}
}
2. 给你一个字符串，以这个字符串中字符出现的频率为权重，构造这个字符串的哈夫曼编码。（题目给的很长，其实意思简洁明了。）
输入样例：
abbcccdddd
输出样例：
1101111111010100000
已经AC代码：
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.priorityqueue;
import java.util.Scanner;
public class Main2 {
static class Tree {
private Node root;
public Node getRoot() {
return root;
}
public void setRoot(Node root) {
this.root = root;
}
}
static class Node implements Comparable<Node> {
private String Strchars = "";
private int Numfrequence = 0;
private Node parent;
private Node leftNode;
private Node rightNode;
@Override
public int compareTo(Node n) {
return Numfrequence - n.Numfrequence;
}
public boolean isLeaf() {
return Strchars.length() == 1;
}
public boolean isRoot() {
return parent == null;
}
public boolean isLeftChild() {
return parent != null && this == parent.leftNode;
}
public int getFrequence() {
return Numfrequence;
}
public void setFrequence(int frequence) {
this.Numfrequence = frequence;
}
public String getChars() {
return Strchars;
}
public void setChars(String chars) {
this.Strchars = chars;
}
public Node getParent() {
return parent;
}
public void setParent(Node parent) {
this.parent = parent;
}
public Node getLeftNode() {
return leftNode;
}
public void setLeftNode(Node leftNode) {
this.leftNode = leftNode;
}
public Node getRightNode() {
return rightNode;
}
public void setRightNode(Node rightNode) {
this.rightNode = rightNode;
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
Map<Character, Integer> charAndNum = statisticsCharNum(input.toCharArray());
String encodedBinariStr = encode2binary(input, charAndNum);
System.out.println(encodedBinariStr);
}
public static Map<Character, Integer> statisticsCharNum(char[] charArray) {
Map<Character, Integer> map = new HashMap<Character, Integer>();
for (char temp : charArray) {
Character Tempchar = new Character(temp);
if (map.containsKey(Tempchar)) {
map.put(Tempchar, map.get(Tempchar) + 1);
} else {
map.put(Tempchar, 1);
}
}
return map;
}
/**
* 构建树
* @param statistics
* @param leafs
* @return
*/
private static Tree buildTree(Map<Character, Integer> statisticsChar2Num, List<Node> leafs) {
Character[] keys_char = statisticsChar2Num.keySet().toArray(new Character[0]);
PriorityQueue<Node> priorityQueue = new PriorityQueue<Node>();
for (Character character : keys_char) {
Node node = new Node();
node.Strchars = character.toString();
node.Numfrequence = statisticsChar2Num.get(character);
priorityQueue.add(node);
leafs.add(node);
}
int sizeNum = priorityQueue.size();
for (int i = 1; i <= sizeNum - 1; i++) {
Node nodeOne = priorityQueue.poll();
Node nodeTwo = priorityQueue.poll();
Node totalNode = new Node();
totalNode.Strchars = nodeOne.Strchars + nodeTwo.Strchars;
totalNode.Numfrequence = nodeOne.Numfrequence + nodeTwo.Numfrequence;
totalNode.leftNode = nodeOne;
totalNode.rightNode = nodeTwo;
nodeOne.parent = totalNode;
nodeTwo.parent = totalNode;
priorityQueue.add(totalNode);
}
Tree tree = new Tree();
tree.root = priorityQueue.poll();
return tree;
}
public static String encode2Binary(String inputStr, Map<Character, Integer> statisticsChar2Num) {
if (inputStr == null || inputStr.equals("")) {
return "";
}
char[] char_Array = inputStr.toCharArray();
List<Node> leaf_Nodes = new ArrayList<Node>();
buildTree(statisticsChar2Num, leaf_Nodes);
Map<Character, String> encod2BinaryInfo = buildEncoding2Binary(leaf_Nodes);
StringBuffer buffer = new StringBuffer();
for (char temp : char_Array) {
Character character = new Character(temp);
buffer.APPend(encod2BinaryInfo.get(character));
}
return buffer.toString();
}
private static Map<Character, String> buildEncoding2Binary(List<Node> leafNodes) {
Map<Character, String> encodewords = new HashMap<Character, String>();
for (Node leafNode : leafNodes) {
Character character = new Character(leafNode.getChars().charAt(0));
String encodeword = "";
Node currentNode = leafNode;
do {
if (currentNode.isLeftChild()) {
encodeword = "0" + encodeword;
} else {
encodeword = "1" + encodeword;
}
currentNode = currentNode.parent;
} while (currentNode.parent != null);
encodewords.put(character, encodeword);
}
return encodewords;
}
}
Reference:
【1】https://blog.csdn.net/kimylrong/article/details/17022319
3. 题目很长，读了好久都没明白题目意思。在此就不列出啦！
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