分享知识-快乐自己：关于 String 小案例

单个字符出现的次数：

/***
     * 验证是否符合拆分条件
     * 
     * @param text
     *            原字符串
     * @param sub
     *            判断条件
     * @return
     */
    public static int count(String text, String sub) {
        int count = 0, start = 0;
        while ((start = text.indexOf(sub, start)) >= 0) {
            start += sub.length();
            count++;
        }
        return count;
    }

一刀切上式（事先补齐）：

@Test
public void demo1() {
    String a = "ab bc,cd,ef gh";
    String result = ""; //保存结果
    a += ","; //事先补齐  ab bc,cd,ef gh,
    for (int i = 0; i < a.length(); i++) {
        char c = a.charAt(i); //获取每一个字符
        if (c == ' ' || c == ',') {
            System.out.print(result);
            result = "";
        } else {
            result += c;
        }
    }
}

一刀切下式（事后修正）：

@Test
public void demo2() {
    String b = "abcdefg"; //结果a,b,c,d,e,f,g
    String result = "";
    for (int i = 0; i < b.length(); i++) {
        result += "," + b.charAt(i);//获取每一个字符
        //,a,b,c,d,e,f,g
    }
    result = result.substring(1);
    System.out.println(result);
}

字符串的常用API  （假设修正法）

@Test
public void demo3() {
    String b = "abcdefgabcd";  //求字符串中是否有重复的字符
    boolean flag = false; // 假设不重复
    for (int i = 0; i < b.length(); i++) {
        char c = b.charAt(i);//获取每一个字符
        if (b.lastIndexOf(c) != i) {
            flag = true; //重复
            break;
        }
    }
}

字符串的常用API  （正则表达式）split

@Test
public void demo4() {
    String  a="a b  c   d e f  g";
    String [] str=a.split(" +"); //a.split(" {1,}");
    for(String  s:str){
        System.out.print(s);
    }
}

字符串替换

@Test
public void demoText1() {
    String str = "a,A,b,c,d,G,h,-Z";
    char[] chars = str.toCharArray();
    for (int i = 0; i < str.length(); i++) {
        int count = (int) chars[i];
        if ((count >= 65 && count <= 90) || (count >= 97 && count <= 127)) {
            System.out.print(chars[i]);
        }
    }
}

字符串中单个字符出现的次数

@Test
public void demoText2() {
    String text = "wefawefaafickdbbjoasdfcweccwerqeasdaajfnblsdbboioe";
    StringBuffer sb = new StringBuffer();
    char[] chars = text.toCharArray();
    Map<String, Integer> map = new HashMap<String, Integer>();

    for (int i = 0; i < chars.length; i++) {
        Pattern pattern = Pattern.compile("(" + chars[i] + ")");//正则表达式 匹配 aa或bb或bb
        Matcher matcher = pattern.matcher(text);
        String t = "";
        while (matcher.find()) {
            t += matcher.group();
        }
        map.put(t.substring(0, 1), Integer.valueOf(t.length()));
    }
    Set<Map.Entry<String, Integer>> entrys = map.entrySet();
    for (Map.Entry<String, Integer> entry : entrys) {
        String key = entry.getKey();
        Integer value = entry.getValue();
        System.out.println(key + "----" + value);
    }
    System.out.println("-----------------------------------------");
    String a = "ab bc,cd,ef gh";
    String result = ""; //保存结果
    a += ","; //事先补齐  ab bc,cd,ef gh,
    for (int i = 0; i < a.length(); i++) {
        char c = a.charAt(i);
        if ((c == ' ') || (c == ',')) {
            System.out.print(result);
            result = "";
        } else {
            result += c;
        }
    }
}

只输出英文字符

@Test
public void demoText3() {
    String str1 = "ab *s-fh*-d-+bc,cd,ef gh";
    char[] chars = str1.toCharArray();
    for (int i = 0; i < str1.length(); i++) {
        int count = (int) chars[i];
        if ((count >= 65 && count >= 90) || (count >= 97 && count >= 127)) {
            System.out.print(chars[i]);
        }
    }
}

任意进制转换首先将3进制转换为10进制，然后再把10进制转换5进制

@Test
public void demoText4() {
    String num = "200110102"; // 3进制的数字转换成5进制的
    int result = 0;
    for (int i = 0; i < num.length(); i++) {
        char c = num.charAt(i);
        result = result * 3 + (c - '0');
    }
    System.out.println(result);  //十进制  13457
    String str = "";
    while (true) {
        if (result == 0) break;
        str += result % 5;
        result = result / 5;
    }
    System.out.println(str);
}

字符串和数字的转换

@Test
public void demoText5() {
    String str = "6789"; //字符串 需要转换成数字  6789
    int num = str.charAt(0) - '0'; // 6
    num = num * 10 + str.charAt(1) - '0';
    num = num * 10 + str.charAt(2) - '0';
    num = num * 10 + str.charAt(3) - '0';
    System.out.println(num);
    System.out.println("------------------------------");
    String str1 = "6789"; //字符串 需要转换成数字  6789
    int result = 0;
    for (int i = 0; i < str.length(); i++) {
        result = result * 10 + str.charAt(i) - '0';
    }
    System.out.println(result);
}

枚举与剪枝 求位数需求:数字自身平方的尾数=自身 （100W以内）

@Test
public void demoText6() {
    for (int a = 0; a <= 1000000; a++) {
        //求a的平方
        int num = a * a;
        if (num % 10 == a || num % 100 == a || num % 1000 == a || num % 10000 == a || num % 100000 == a || num % 1000000 == a) {
            System.out.println("数字本身是=》" + a + "数字的平方是=》" + num);
        }
    }
}

鸡兔同笼

/**
 * 鸡兔同笼
 * 鸡兔共50头，脚共120只！
 * x+y=50;
 * 2x+4y=120;
 */
@Test
public void demoText7() {
    for (int x = 0; x <= 50; x++) {
        int y = 50 - x; //得到兔子的数量
        if (2 * x + 4 * y == 120) {
            System.out.println(x);
            System.out.print(y);
        }
    }
}

判断某个字符串是否满足我们定义的正则表达式

@Test
public void demoText8() {
    String a = "ABC01234";
    boolean matches = a.matches("[A-Z]{1,3}[0-9]{3,5}");//返回boolean
    System.out.println(matches);
}

字符串的常用API （正则表达式）replaceAll()在小括号里面的内容我们称之为子组，用$获取子组内容

@Test
public void demoText9() {
    String a = "asasasa 2018-07-16 sasa sasas sasa";
    //实现的效果是asasasa 07/16/2018年 sasa sasas sasa
    a = a.replaceAll("([0-9]{4})-([0-9]{2})-([0-9]{2})", "$2/$3/$1 年");
    System.out.print(a);
}

文字正序输出，并显示字母出现次数

public static void main(String[] args) {
    String str = "服微端云入加迎欢|welcome join us !";
    int flag=0;
    String str1=str.substring(0, 8);
    String str2=str.substring(9);
    StringBuffer sb1=new StringBuffer(str1);
    sb1.reverse();
    String newStr=sb1.append(str2).toString();
    System.out.println(newStr);
    for (int i = 0; i < str2.length(); i++) {
        char letter=str2.charAt(i);
        if (String.valueOf(letter).matches("^[a-zA-Z]$")) {
            flag++;
        }
    }
    System.out.println("英文字符有："+flag+"个");
}