Question

长篇小说初学者程序员在这里用java练习。现在我正在使用两个数组，我想知道int arrayA（7,14,21,28）是否连续到int ArrayB意味着arrayB在数组中连续有（7,14,21,28）。如果是boolean则返回true，否则它将返回false。这是我的代码示例。

 public class Harrison7aTest
 {
 public static void main(String [] args)
 {
  int[] arrayA = {7,14,21,28};
  int[] arrayB = {1,3,5,7,14,21,28,32};

 boolean result = false;

  for(int A = 0; A < arrayA.length - 1; A++)
  {
     for(int B = 0; B < arrayB.length - 1; B++)
     {


     }


  }

}
}

Answer 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

Answer 2

如果您想使用该方法，您可能会在生产中用作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上进行一次迭代，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));
}

Answer 3

     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);

更新了一个：

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;

}

}

参见上面的例子。

Answer 4

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

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);
 }

Java比较两个数组并检查First数组是否是Second的连续数

4 个答案: