面試（編程題）

時間 2019-12-08

原文原文鏈接

1. 在不使用 StringBuffer 的前提下，怎麼反轉一個字符串？解決方案html

思路：將字符串轉成字符數組，倒敘遍歷，而後拼接成字符串。java

public class StringReverseExample {
    public static void main(String args[]) {

        //quick wasy to reverse String in Java - Use StringBuffer
        String word = "HelloWorld";
        String reverse = new StringBuffer(word).reverse().toString();
        System.out.printf(" original String : %s , reversed String %s  %n", word, reverse);

        //another quick to reverse String in Java - use StringBuilder
        word = "WakeUp";
        reverse = new StringBuilder(word).reverse().toString();
        System.out.printf(" original String : %s , reversed String %s %n", word, reverse);

        //one way to reverse String without using StringBuffer or StringBuilder is writing
        //own utility method
        word = "Band";
        reverse = reverse(word);
        System.out.printf(" original String : %s , reversed String %s %n", word, reverse);
    }

    public static String reverse(String source) {
        if (source == null || source.isEmpty()) {
            return source;
        }
        String reverse = "";
        for (int i = source.length() - 1; i >= 0; i--) {
            reverse = reverse + source.charAt(i);
        }

        return reverse;
    }
}

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2. Java 中，怎麼獲取一個文件中單詞出現的最高頻率？解決方案git

思路：將文本中的字符分詞，而後使用 HashMap 的存儲，key 存儲單詞，value 存儲出現次數。正則表達式

public class Problem {
    public static void main(String args[]) {
        Map<String, Integer> wordMap = buildWordMap("C:/temp/words.txt");
        List<Map.Entry<String, Integer>> list = sortByValueInDecreasingOrder(wordMap);
        System.out.println("List of repeated word from file and their count");
        for (Map.Entry<String, Integer> entry : list) {
            if (entry.getValue() > 1) {
                System.out.println(entry.getKey() + " => " + entry.getValue());
            }
        }
    }

    public static Map<String, Integer> buildWordMap(String fileName) {
        // Using diamond operator for clean code
        Map<String, Integer> wordMap = new HashMap<>();
        // Using try-with-resource statement for automatic resource management
        try (FileInputStream fis = new FileInputStream(fileName);
             DataInputStream dis = new DataInputStream(fis);
             BufferedReader br = new BufferedReader(new InputStreamReader(dis))) {
            // words are separated by whitespace
            Pattern pattern = Pattern.compile("\\s+");
            String line = null;
            while ((line = br.readLine()) != null) {
                // do this if case sensitivity is not required i.e. Java = java
                line = line.toLowerCase();
                String[] words = pattern.split(line);
                for (String word : words) {
                    if (wordMap.containsKey(word)) {
                        wordMap.put(word, (wordMap.get(word) + 1));
                    } else {
                        wordMap.put(word, 1);
                    }
                }
            }
        } catch (IOException ioex) {
            ioex.printStackTrace();
        }
        return wordMap;
    }

    public static List<Map.Entry<String, Integer>> sortByValueInDecreasingOrder(Map<String, Integer> wordMap) {
        Set<Map.Entry<String, Integer>> entries = wordMap.entrySet();
        List<Map.Entry<String, Integer>> list = new ArrayList<>(entries);
        list.sort((o1, o2) -> (o2.getValue()).compareTo(o1.getValue()));
        return list;
    }
}

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3. 在沒有使用臨時變量的狀況如何交換兩個整數變量的值？數組

思路：本身構造第三方變量。緩存

int a = 5, b = 8;

// 方法一：加減法
a = a + b; // 本身造一個臨時變量
b = a - b;
a = a - b;
// 方法二：異或法
a = a ^ b; // 0101 ^ 1000 = 1100
b = b ^ a; // 1000 ^ (0101 ^ 1000) = 1000 ^ 1000 ^ 0101 = 0000 ^ 0101 = 0101
a = a ^ b; // (0101 ^ 1000) ^ 0101 = 0101 ^ 0101 ^ 1000 = 0000 ^ 1000 = 1000

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4. Java 中如何將字符串轉換爲整數？app

思路：①首先校驗字符串；②判斷數字開頭是正號仍是負號，設置一個標誌位，正號爲 1，負號爲 -1；③判斷字符是否在 0 到 9 以內，是轉數字，不然跳出。ide

public static int stringToInt(String str) {
    int index = 0;
    int sign = 1;
    int digit = 0;
    int total = 0;
    char ch;

    // 1.1判斷字符串是否爲空。
    if (str.length() == 0) {
        return 0;
    }

    // 1.2判斷字符串中是否含有空格，若是有就跳過。
    while (str.charAt(index) == ' ' && index < str.length()) {
        index++;
    }

    // 2.判斷字符串爲正仍是負。
    if (str.charAt(index) == '+' || str.charAt(index) == '-') {
        if (str.charAt(index) == '+')
            sign = 1;
        else {
            sign = -1;
        }
        index++;
    }

    // 3.進行處理。
    while (index < str.length()) {
        ch = str.charAt(index);
        if (ch < '0' || ch > '9')
            break;
        digit = ch - '0';

        //校驗後續 total = total * 10 + digit; 操做以後數字的絕對值的大小，若是超出 MAX_VALUE，則根據負號標誌位進行相應的賦值。
        if (Integer.MAX_VALUE / 10 < total || Integer.MAX_VALUE / 10 == total && Integer.MAX_VALUE % 10 < digit)
            return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;

        total = total * 10 + digit;
        index++;

    }
    return sign * total;
}

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5. 給定一個 txt 文件，如何獲得某字符串出現的次數。ui

思路：使用正則表達式。this

public static int count() throws IOException {
    //讀入文本(字符)，用FileReader->BufferedReader
    //讀入輸入(字節)，用FileInputStream->InputStreamReader->BufferedReader
    String path = Test.class.getResource("/1.txt").getPath();
    BufferedReader br = new BufferedReader(new FileReader(path));
    StringBuilder sb = new StringBuilder();

    while (true) {
        String str = br.readLine();
        if (str == null) {
            break;
        }
        sb.append(str);
    }
    Pattern p = Pattern.compile("mobnet");
    Matcher m = p.matcher(sb);
    int count = 0;
    while (m.find()) {
        count++;
    }
    return count;
}