Java比较两个数组并检查First数组是否是Second的连续数

4 回答

• 1

  您可以将它们转换为字符串并使用 str.contains() 方法

  String strArrA = Arrays.toString(arrayA);
  String strArrB = Arrays.toString(arrayB);
  
  //to strip square brackets that's comes with Arrays.toString
  //for ex: Arrays.toString(arrayA); returns "[7, 14, 21, 28]"
  //we want to convert it to "7, 14, 21, 28"
  strArrA = strArrA.substring(1, strArrA.length()-1);
  
  if (strArrB.contains(strArrA)) {
      System.out.println("true");
  } else {
      System.out.println("false");
  }
  

  DEMO

  回复于 2024-05-06T00:41:35+08:00
• 3

  如果您想使用该方法，您可能会在 生产环境 中用作Java工程师，然后检查@RC给出的重复链接 . 或@Raman的答案 . 出于作业的目的，如果您只想使用单个循环回答问题，请考虑我的答案 . 您可以在 arrayB 上迭代一次，并检查 arrayA 中包含的数字序列是否发生，而不会中断 .

  public static boolean containsConsecutive(int[] a, int[] b) {
      int aIndex = 0;
      for (int i=0; i < arrayB.length; i++) {
          if (aIndex != 0 && arrayB[i] != arrayA[aIndex]) {
              break;
          }
          else if (arrayB[i] == arrayA[aIndex]) {
              ++aIndex;
          }
      }
  
      return aIndex == arrayA.length;
  }
  
  public static void main(String[] args) {
      int[] a = {7,14,21,28,32};
      int[] b = {1,3,5,7,14,21,28,32};
      // true
      System.out.println(containsConsecutive(a, b));
      a = {7,14,21,28,32};
      b = {1,3,5,7,8,9,10,14,21,28,32};
      // false - sequence not in order
      System.out.println(containsConsecutive(a, b));
      a = {7,14,21,28,32,35};
      b = {1,3,5,7,14,21,28,32};
      // false - entire sequence in a not contained within b
      System.out.println(containsConsecutive(a, b));
  }
  

  回复于 2024-05-06T00:41:35+08:00
• 0

  int[] arrayA = {7,14,21,28};
       int[] arrayB = {1,3,5,7,14,21,28,32};
       boolean result=false;
  
       for(int i=0;i<arrayB.length;i++){
  
           if(arrayA[0] == arrayB[i]){
               for(int j=0;j<arrayA.length;j++){
                   if(arrayA[j] == arrayB[i+j]){
                       result=true;
                   }
                   else{
                       result=false;
                   }
               }
           }
       }
      System.out.println(result);
  

  更新一个：

  public class Test {
  
  public static void main(String[] args) {
  
      int[] arrayA = { 7, 14, 21, 28 };
      int[] arrayB = { 1, 3, 5, 7, 14, 21, 28, 32, 7 };
      boolean output = Test.appearsConsecutive(arrayA,arrayB);
      System.out.println(output);
  }
  
  public static boolean appearsConsecutive(int[] arrayA, int[] arrayB) {
  
      boolean result = false;
  
      for (int i = 0; i < arrayB.length; i++) {
  
          if (arrayA[0] == arrayB[i]) {
              for (int j = 0; j < arrayA.length; j++) {
                  if (arrayA[j] == arrayB[i + j]) {
                      result = true;
                      break;
                  } else {
                      result = false;
                  }
              }
          }
      }
      System.out.println(result);
      return result;
  
  }
  
  }
  

  见上例 .

  回复于 2024-05-06T00:41:35+08:00
• 1

  如果有人需要算法答案....（我已经颠倒了arrayA和arrayB赋值） .

  public static void main (String[] args) throws java.lang.Exception
  {
      // your code goes here
      int[] arrayB = {7,14,21,28};
      int[] arrayA = {1,3,5,7,14,21,28,32};
  
      boolean result = false;
  
      for(int A = 0; A < arrayA.length; A++)
      {
           if( arrayA[A] != arrayB[0]) continue;
           //else ... 
           for(int B = 0; B < arrayB.length; B++)
           {
                      if(arrayA[A] != arrayB[B] || A>=arrayA.length) 
                          break; 
                      if(B+1 == arrayB.length) 
                          {
                              result = true;
                              break;
                          }
                      A++;
           }
          if(result) 
              break;
  
        }
        System.out.println("Contains :"+ result);
   }
  

  回复于 2024-05-06T00:41:35+08:00