面试题及个人的解法

Posted 2022-01-13 _Slience_

这段时间在找工作，有一家在面试的时候的编程题我写了伪代码，现在以完整的代码写出来，有参考网上，以下是题目

1

编写一个截取字符串的函数，输入为一个字符串和字节数， 输出为按字节截取的字符串。但是要保证汉字不被截半个，如"我ABC"4，应该截为"我AB"，输入“我ABC汉DEF", 6，应该输出为’我ABC"而不是"我ABC"+汉的半个

    public void testSpiltStr() 
        /*String str = "我ABC";
        int num = 4;*/

        String str = "我ABC汉DEF";
        int num = 6;
        
        char[] chars = str.toCharArray();
        int index = 0;
        for(int i = 0; i < str.length(); i++) 
            if(isChinese(chars[i])) 
                index += 2;
             else 
                index++;
            
            if(index > num) 
                break;
            
            System.out.print(chars[i]);
        
    

    private static boolean isChinese(char c) 
        Character.UnicodeBlock ub = Character.UnicodeBlock.of(c);
        if (ub == Character.UnicodeBlock.CJK_UNIFIED_IDEOGRAPHS || ub == Character.UnicodeBlock.CJK_COMPATIBILITY_IDEOGRAPHS
                || ub == Character.UnicodeBlock.CJK_UNIFIED_IDEOGRAPHS_EXTENSION_A || ub == Character.UnicodeBlock.CJK_UNIFIED_IDEOGRAPHS_EXTENSION_B
                || ub == Character.UnicodeBlock.CJK_SYMBOLS_AND_PUNCTUATION || ub == Character.UnicodeBlock.HALFWIDTH_AND_FULLWIDTH_FORMS
                || ub == Character.UnicodeBlock.GENERAL_PUNCTUATION) 
            return true;
        
        return false;
    

运行结果

2

试用递归的方法写一下计算菲波那契数列的通项.f(n)，已知f1=1,f2=1，以后每项都是前两项的和

    public void testFibonacci() 
        int num = 23;

        System.out.println(fibonacci(num));
    
    
    public int fibonacci(int n) 
        if(n == 1 || n == 2) 
            return 1;
         else 
            return fibonacci(n-1) + fibonacci(n-2);
        
    

3

给你一组字符如1, 3, 4, 7, 2, 1, 1, 5, 2让你输出里面出现次数最多且数值最大的一个，出现了几次

    public void testNumCount()
        Integer[] nums = 1, 3, 4, 7, 2, 1, 1, 5, 2;
        
        ArrayList<Integer> numList = new ArrayList<Integer>(Arrays.asList(nums));
        ArrayList tempList = new ArrayList<Integer>(new LinkedHashSet<Integer>(numList));
        for(int i = 0; i < tempList.size(); i++) 
            System.out.println(tempList.get(i) + "出现次数:" + Collections.frequency(numList, tempList.get(i)));
        
    

这里使用LinkedHashSet是为了去重并且保持原有的顺序