Java。如何從兩個字符串數組或列表中刪除相同的字符串對象

這是我的第一個問題，它是關於Java的。我想實現以下邏輯：Java。如何從兩個字符串數組或列表中刪除相同的字符串對象

我有兩個字符串數組（或字符串的字符串列表）。有一個字符串排列（asu） - M1，M2，M3 ...以及一串字符串（rzs） - M1，M2，M3及其所有可能的組合。每個元素（asu）（例如M1）需要找到包含（例如M1）的（rzs）（M1, M1M2，...）中的元素。例如：從（asu）中取出M1，將開始在（rzs）中搜索重複（包含）。我們在（rzs）找到M1M2，它包含M1。之後，我們應該從數組（列表）中刪除這兩個元素。而我，我的英語技能對不起^^

String[] asu = { "M1", "M1", "M1", "M3", "M4", "M5", "M1", "M1", "M1", "M4", "M5", "M5" }; 
String[] rzs = { "M1", "M2", "M3", "M4", "M5", "M1M2", "M1M3", "M1M4", "M1M5", "M2M3", "M2M4", "M2M5", "M3M4", "M3M5", "M4M5", "M1M2M3", "M1M2M4", 
     "M1M2M5", "M1M3M4", "M1M3M4", "M1M4M5", "M2M4", "M2M5" }; 

public static void main(final String[] args) { 
    work bebebe = new work(); 
    bebebe.mywork(); 
} 

public void mywork() { 

    System.out.println(Arrays.deepToString(rzs)); 
    System.out.println(Arrays.deepToString(asu)); 
    for (int i = 0; i < asu.length; i++) { 
     System.out.println("Итерация: " + i); 
     for (int j = 0; j < rzs.length; j++) { 
      if (asu[i].matches(rzs[j].toString())) { 
       System.out.println(i + " элемент (" + asu[i] + ") в ASU равен " + j + " элементу (" + rzs[j] + ") в RZS"); 
       asu[i] = ""; 
       rzs[j] = ""; 
      } 
     } 
    } 
}

結果不會刪除的子串項目。不滿足邏輯。我會感謝您的建議。

來源

2013-08-05 Eldar Nezametdinov

如果你會使用列表你有一個好很多：刪除不會問你拿陣列的休息回來，和使用名單意味着你的代碼

較少的邏輯

List<string> asu = Arrays.asList("M1","M1","M1","M3","M4","M5","M1","M1","M1","M4","M5","M5"); 
List<string> rzs = Arrays.asList("M1","M2","M3","M4","M5", 
"M1M2","M1M3","M1M4","M1M5","M2M3","M2M4","M2M5","M3M4","M3M5","M4M5" 
,"M1M2M3","M1M2M4","M1M2M5","M1M3M4","M1M3M4","M1M4M5","M2M4","M2M5"); 
public static void main(String[] args) { 
work bebebe = new work(); 
bebebe.mywork(); 
} 

public static void mywork() { 
    ArrayList<String> tmp1 = new ArrayList<String>(); 
    ArrayList<String> tmp2 = new ArrayList<String>(); 
    System.out.println((rzs)); 
    System.out.println((asu)); 
    for (String curr : asu){ 
    for (String currRzs : rzs){ 
     if (currRzs.contains(curr)) { 
      System.out.println(" item("+curr+") in ASU found contained in ("+currRzs+") in RZS"); 

     if(tmp1.contains(curr) == false) 
     tmp1.add(curr); 

     if(tmp2.contains(currRzs) == false) 
     tmp2.add(currRzs); 
     } 
     } 
    } 

    for (String curr : tmp1){ 
     asu.remove(curr); 
    } 

    for (String curr : tmp2){ 
     rzs.remove(curr); 
    } 
}

來源

2013-08-05 06:54:04

你的代碼很美觀，但你的解決方案也不滿足邏輯。正如我之前所說，如果我們發現M1M2（rzs），我們應該將他與M1比較。如果M1是M1M2的子串，我們應該刪除兩者。你的代碼不會這樣做。「包含」語句只比較像M1 = M1這樣的對象，但不包含對象的字符串; –

@EldarNezametdinov包含不等於。你可以將字符串框到你自己的類，並覆蓋等於做indexOf。你想要一個例子嗎？ –

如果對你來說不難，因爲我幾天前纔開始學習Java ......我不知道這種情況怎麼做纔對。 –

Java。如何從兩個字符串數組或列表中刪除相同的字符串對象

回答

相關問題