java - 检查数组是否有连续的元素？

Question

大家好，我想问一下，我是否想检查数组是否有 4 个连续元素，就像这样 {1,2,4,4,4,4,3,5} 它将返回 true 但如果数组是 {1,2,4,4,4,2,4,2} 返回 false请使用 Java 提供帮助:)}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

独特的解决方案，供引用:

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。当数字序列在一行中出现四次(以逗号分隔)时，即匹配。这是分解后的样子:

> 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}