Question

大家好我想问一些问题，如果我想检查数组是否有4个连续的元素，或者不是这个{1,2,4,4,4,4,3,5}它会返回true但是如果数组是{1,2,4,4,4,2,4,2}返回false请帮助Java :)}这是我的代码：

public class arraysearch {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int arr [] ={1,2,4,4,4,4,5,7,10};
        int start = arr[0];
        int count = 0;
        int sum=0;
        for(int i = 1;i<arr.length-1;i++)
        {
            if(start==4)
            {
                count++;
            }
            else
            {
                //count=0;
                for(int j=i+1;j<arr.length-1;j++){
                    if(arr[i]==arr[j]&&arr[j]==4)
                    {
                        count++;
                        //continue;
                    }
                }
            }
        }
        if(count == 4)
        {
            System.out.println("it has 4 elements");
        }
        else
            System.out.println("it hasn't 4 elements");
    }

}

Answer 1

一个独特的解决方案，供参考：

int arr [] ={1,2,4,4,4,4,5,7,10};
for(int o : arr)
  buffer.append(",").append(o);
boolean match = buffer.toString().matches("(\\d+)(?:,\\1){3}");
System.out.println("It "+(match ? "has" : "hasn't")+ " 4 elements");

您可以重复使用该模式以提高效率：

// Class
private static final Pattern CHAIN=Pattern.compile("(\\d+)(?:,\\1){3}");

// Method
int arr [] ={1,2,4,4,4,4,5,7,10};
for(int o : arr)
  buffer.append(",").append(o);
System.out.println("It "+(CHAIN.matcher(buffer).matches() ? "has" : "hasn't")+ " 4 elements");

让我在这里解释一下这个魔法：

(\d+)(?:,\1){3}是正则表达式。当一行数字连续四次出现时，它与逗号分开。以下是分解后的样子：

> Capturing Group "\d+" one or more number digits
|     > Non- Capturing Group: ",\1" a comma and a captured group
|     |      > The second group matches 3 times
|     |      |
(\d+)(?:,\1){3}

Answer 2

您的代码不会区分连续元素和分隔元素。你可以做什么（略微修改）重置计数。

count = 1;
for(int i=0; i<arr.size-1; i++){
  if(arr[i+1] == arr[i]){
     count++;
  }
  else{
      count=1;
  }
  if(count == 4){
      return true;
  }
}

此代码通过检查下一个元素是否与当前元素相同来迭代，如果是，则向计数添加一个元素。如果不是，则重置下一个链的计数。这适用于任何一系列数字，并且总是存储最长的数字，所以我建议检查它是否包含一系列至少4个元素。

Answer 3

试试这个：

      public class SumDemo {
         public static void main(String[] args) {
           // TODO Auto-generated method stub
            int arr[] = {
            1, 2, 4, 4, 4, 4, 5, 7, 10
          };

          int count = 0;

    for (int i = 0; i < arr.length - 1; i++) {
        if (arr[i] != arr[i + 1])
            continue;
        else
            count++;

    }
    if (count >= 3) { // three comparision found true, means the array has 4
                      // simliar continuous elements
        System.out.println("it has " + count + "elements");
    } else
        System.out.println("it has " + count + "elements");

}

}

如果该解决方案具有多个连续的元素，则该解决方案将无效。例如{1,2,2,4,4,4,4,5,7,10}

Answer 4

好吧，你的代码似乎有些搞砸了。您询问您的代码是否有4个连续的元素，但在您的代码中，您似乎正在测试它是否连续4个4。假设您的数组至少包含1个元素：

int[] arr = new int[]{1, 2, 4, 4, 4, 2, 4, 2};

//current consecutive sector
int currentLength = 1;
int currentValue = arr[0];

//longest consecutive sector so far
int maxLength = 1;
int maxValue = currentValue;

for (int i = 1; i < arr.length; i++) {
    if (currentValue == arr[i]) {
        //hooray, current consecutive sector is getting larger
        currentLength++;
    } else {
        //oh no, current consecutive sector ends and starts a new one
        if (currentLength > maxLength) {
            maxLength = currentLength;
            maxValue = currentValue;
        }
        currentLength = 1;
        currentValue = arr[i];
    }
}

//finally, check if the last sector isn't the longest
if (currentLength > maxLength) {
    maxLength = currentLength;
    maxValue = currentValue;
}

System.out.format("Longest consecutive sector contained %d times number %d.\n",
        maxLength, maxValue);

检查数组是否有连续的元素？

4 个答案: