请用 Java 写一个 String 处理方法）

请用 Java 写一个 String 处理方法，将输入字符串中的中文与英文、数字之间使用空格隔开，并针对特殊符号（括号和逗号）

package com.fufeiwen.test;

import java.util.regex.Pattern;

public class Test {

public static void main(String args[]) {
    System.out.println(getStr("我是中国ren，nishi谁？"));
}

public static String getStr(String str) {
    String curStr;
    int preType = -1;
    StringBuffer sb = new StringBuffer();
    for (int i = 0; i < str.length(); i++) {
        curStr = str.substring(i, i + 1);
        int curType = getType(curStr);
        if (i>0&&curType != preType) {
            sb.append(" ");
        }
        sb.append(curStr);
        preType=curType;
    }
    return sb.toString();
}

public static int getType(String str) {
    String chinese = "^[\u4e00-\u9fa5]{0,}$";
    String number = "^[0-9]*$";
    String letter = "^[A-Za-z]+$";
    int num = 0;//其他
    if (Pattern.matches(chinese, str)) {
        num = 1;//中文
    } else if (Pattern.matches(number, str)) {
        num = 2;//数字
    } else if (Pattern.matches(letter, str)) {
        num = 3;//字母
    }
    return num;
}

}

以上是大体思路，没考虑性能、细节。

package com.test.spring.test;

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Test {
public static void main(String [] arg){
String str = "张三1234589*李四ABCabc王五2";
System.out.println(setStr(str));
}

public static String setStr(String str){
String value = "";
char [] c = str.toCharArray();
value += c[0];
for (int i = 1; i < c.length; i++) {
if( (isLetter(c[i]) == true && isNumber(c[i-1]) ==true) ||
(isLetter(c[i]) == true && isChinese(c[i-1]) ==true) ||
(isLetter(c[i]) == true && isSpecialCharacter(c[i-1]+"") ==true) ||
(isNumber(c[i]) == true && isLetter(c[i-1]) ==true) ||
(isNumber(c[i]) == true && isChinese(c[i-1]) ==true) ||
(isNumber(c[i]) == true && isSpecialCharacter(c[i-1]+"") ==true) ||
(isChinese(c[i]) == true && isLetter(c[i-1]) ==true) ||
(isChinese(c[i]) == true && isNumber(c[i-1]) ==true) ||
(isChinese(c[i]) == true && isSpecialCharacter(c[i-1]+"") ==true) ||
(isSpecialCharacter(c[i]+"") == true && isLetter(c[i-1]) ==true) ||
(isSpecialCharacter(c[i]+"") == true && isNumber(c[i-1]) ==true) ||
(isSpecialCharacter(c[i]+"") == true && isChinese(c[i-1]) ==true)

)
{
value = (value + " " + c[i]);
}else{
value = (value + c[i]);
}
}
return value;
}
/**
* 判断是否为字母
* @param c
* @return
/
public static boolean isLetter(char c){
if((c >= 'a' && c <= 'z') || ( c >= 'A' && c <= 'Z') )
return true;
return false;
}
/*
* 判断是否为数字
* @param c
* @return
/
public static boolean isNumber(char c){
if( c >= '0' && c <= '9' )
return true;
return false;
}
/*
* 根据Unicode编码完美的判断中文汉字和符号
* @param c
* @return
/
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;
}
/*
* 判断是否为特殊字符

* @param 字符
* @return
* @throws
/
public static boolean isSpecialCharacter (String c) {

// 正则表达式判断是否存在特殊字符

String regEx="[`~!@#$%^&()+=|{}':;',\[\].<>/?~！@#￥%……&*（）——+|{}【】‘；：”“’。，、？]";

Pattern p = Pattern.compile(regEx);

Matcher m = p.matcher(c);

if(m.find())
return true;
return false;
}
}
大体实现应该是这样的

因为我没看懂 针对特殊符号（括号和逗号）要干嘛，所以我只针对另外三种写的代码：

public static String getStr(String str) {
        String chinese = "([\u4e00-\u9fa5]+)";
        String number = "([0-9]+)";
        String letter = "([A-Za-z]+)";
        String[] regexs = {
                chinese + number,
                chinese + letter,
                number + chinese,
                number + letter,
                letter + chinese,
                letter + number
        };
        for (String regex : regexs) {
            str = str.replaceAll(regex, "$1 $2");
        }
        return str;
    }