如何整理只包含字母的數組？

Question

有2個數組。 代碼[]和文本[]

code[0] = 0, text[0] = "Monkey333banana" // has numbers in text 
code[1] = 100, text[1] = "Dog" 
code[2] = 200, text[2] = "Cat117" // has numbers in text 
code[3] = 300, text[3] = "Pig" 

我想拿起文本的數組[] 僅由字母。 沒有號碼。

code[1] = 100, text[1] = "Dog" 
code[3] = 300, text[3] = "Pig" 

而我想保存另一個新的數組。

newCode[0] = 100, text[0] = "Dog" 
newCode[1] = 300, newText[1] = "Pig" 

示例代碼：

String[] code = new String[4]; 
int[] text = new int[4]; 

code[0] = 0; 
code[1] = 100; 
code[2] = 200; 
code[3] = 300; 

text[0] = "Monkey333banana"; 
text[1] = "Dog"; 
text[2] = "Cat117"; 
text[3] = "Pig"; 

// *** MAGIC **** // 
// *** MAGIC **** // 
// *** MAGIC **** // 

for (int i=0; i<(proper size); i++) 
System.out.println(newCode[i]+" : "+newText[i]); 

結果：

• 100：狗;
• 300：豬;

我該怎麼做這個魔法？請幫幫我。謝謝！

對不起......對行爲輕率，我寫了magic也許有人認爲這是不好的，對不起......

Answer 1

我希望這有助於提供：兩個數組是相同的

• 大小...

  public static void main (String[] args) { 
  
      ArrayList<Integer> list = new ArrayList<Integer>(); // store code[] value 
      ArrayList<String> listS = new ArrayList<String>(); // store string w/o numbers 
  
      int code[] = {0, 100, 200, 300}; //example 
      String text[] = {"M12", "Dog", "117Cat", "Pig"}; //example 
  
      for (int i = 0; i < text.length; i++) { 
  
       if (! text[i].matches(".*\\d+.*")) { // VERY IMP REGEX TO TEST IF STRING CONTAINS ANY NUMBER 
        list.add(i); // or add(code[i]); 
        listS.add(text[i]); 
       } 
      } 
  
      int newCode[] = new int[list.size()]; 
      int idx = 0; 
      for (int x : list) { 
       newCode[idx++] = x; 
      } 
  
      String newText[] = listS.toArray(new String[list.size()]); 
  
      for (int i=0; i < newCode.length; i++) { 
       System.out.println(newCode[i] + " : " + newText[i]); 
      } 
  } 
  

• 而不是寫magic投入一些精力/研究張貼問題

  前

Answer 2

那魔：

for(int i=0;i<text.length;i++) 
    textvalid[i]=true; //saying that all the strings are valid. We will mark them as unvalid later 
k=0; //a counter 
for(int i=0;i<text.length;i++) { //every string in the array 
    for(int j=0, char[] text1 = text.toCharArray();j<text1.length;j++)  //every character in the string. text1 is the char[] equivallent of text[i] 
     if(text1[j]<'A' || (text1[j] >'Z' && text1[j]<'a') || text1[j] <='z') //if it is not a capital letter nor a lowercase one 
      textvalid[i] = false; //mark it as a npn-valid string 
    if(textvalid[i]==true){ //if it is valid, add it to the newtext array 
     newText[k++] = text[i]; 
     newCode[k] = code[i]; 
    } 
} 

希望它能幫助。 :)

Answer 3

這裏我發佈一條提示，進一步它是你的責任，自己動手

for(char i = 'a'; i <= 'z'; i++){ 
System.out.println(i+" "); 
} 

Answer 4

String[] newText = new String[4]; 
    int[] newCode = new int[4]; 
    int j=0; 
    for(int i=0;i<text.length;++i){ 
     if(text[i].matches("\\A[a-zA-Z]+\\z")){ 
      newCode[j]=code[i]; 
      newText[j++]=text[i]; 
     } 
    } 

    for (int i=0; i<j; i++) 
     System.out.println(newCode[i]+" : "+newText[i]);