2019 面试题

题目1：给一段字符串在不用其他数据结构或者API的情况下，反转字符串。尽可能考虑其性能问题

String str = "abc146d3e2f"; 

　　
　　倒着取字符串
　　private static void reverse1(String str) {
        char [] ch = new char[str.length()];
        for(int i=str.length()-1,k=0;i>=0;i--,k++) {
            ch[k] = str.charAt(i);
        }
        System.out.println(ch);
    }
　　//循环次数减半。
    private static void reverse(String str) {
        char [] ch = str.toCharArray();
        int len = ch.length % 2 == 0 ? ch.length / 2: ch.length / 2 +1;
        int max = ch.length -1 ;
        for (int i = 0; i < len; i++) {
            char temp = ch[max];
            ch[max--] = str.charAt(i);
            ch[i] = temp;
        }
        System.out.println(ch);
    }

题目2：给出一段字符串，取出数字。

String str = "abc146d3e2f"; 

　　
　　　　//利用正则
　　private static void filterLetter1(String str1) {
        Set<Character> set = new TreeSet<>();
        Pattern pattern = Pattern.compile("[0-9]*");
        Matcher isNum = null;
        for (int i = 0; i < str1.length(); i++) {
            isNum = pattern.matcher(str1.substring(i, i + 1));
            if (isNum.matches()) {
                set.add(str1.charAt(i));
            }
        }

        System.out.println(set);
    }
　　//利用ascii码特征过滤
    private static void filterLetter(String str1) {
        char[] arr = str1.toCharArray();
        StringBuffer sb = new StringBuffer();
        Set<Character> set = new TreeSet<>();
        for (int i = 0; i < arr.length; i++) {
            int num = (int) arr[i];
            if (num < 97) {
                sb.append((char) num);
            }
        }

        System.out.println(set);
    }

题目3：将一组数组完全打乱

int [] arr = new int []{1,2,3,4,5,6,7,8}

　　private static void scatter(int[] arr) {
        int count = 0;
        while (count < 8) {
            int random = (int) (Math.random() * 8);
            if (arr[random] > 0) {
                System.out.print(arr[random]);
                arr[random] = 0;
                count++;
            }
        }
    }