检查数组中的所有重复元素是否相邻

给定一个数组arr[] 。任务是检查arr[]中的重复元素是否连续。

例子

Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: Yes
Explanation: There is no duplicate element in arr[] so there is no need to check anything and answer is Yes.

Input: arr[] = {1, 2, 2, 4}
Output: Yes
Explanation: 2 is occurring 2 times and it is contiguous. Hence, answer is Yes.

Input: arr[] = {1, 2, 3, 2, 4}
Output: No
Explanation: There is a gap between 2’s and 3 is between two 2’s. Therefore, the answer is No.

编程需要懂一点英语

方法：这个问题可以通过使用HashMaps来解决。请按照以下步骤解决给定的问题。

使用地图存储访问过的元素。
首先，在地图上标记第一个元素。
遍历数组arr[]从1到N-1 。其中N是arr[]的大小。
如果当前元素与前一个元素匹配，则意味着有一个元素重复循环，因此只需继续循环即可。
如果当前元素已在地图中标记，则返回“否” 。
标记地图中的当前元素。
如果函数到达这里意味着有所有连续的元素，所以返回“是” 。

下面是上述方法的实现。

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check whether duplicate
// elements in array arr[] are contiguous or not
string checkContiguous(int* arr, int& n)
{
    int i;
 
    // Map to keep track of elements
    unordered_map visited;
 
    visited.clear();
    visited.insert({ arr[0], 1 });
 
    for (i = 1; i < n; i++) {
        if (arr[i] == arr[i - 1])
            continue;
        else if (visited[arr[i]])
            return "No";
        visited[arr[i]] = 1;
    }
 
    return "Yes";
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 4, 5, 5, 3, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << checkContiguous(arr, N);
 
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG {
 
  // Function to check whether duplicate
  // elements in array arr[] are contiguous or not
  static String checkContiguous(int[] arr, int n)
  {
    int i;
 
    // Map to keep track of elements
    int[] visited = new int[n];
 
    for (i = 1; i < n; i++) {
      if (arr[i] == arr[i - 1])
        continue;
      else if (visited[arr[i]] == 0)
        return "No";
      visited[arr[i]] = 1;
    }
 
    return "Yes";
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    int arr[] = { 2, 4, 5, 5, 3, 5 };
    int N = arr.length;
 
    // Function Call
    System.out.println(checkContiguous(arr, N));
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python program for the above approach
 
# Function to check whether duplicate
# elements in array arr[] are contiguous or not
def checkContiguous  (arr, n):
    i = None
 
    # Map to keep track of elements
    visited = [0] * n;
 
 
    for i in range(1, n):
        if (arr[i] == arr[i - 1]):
            continue;
        elif (visited[arr[i]] == 0):
            return "No";
        visited[arr[i]] = 1;
     
 
    return "Yes";
 
 
# Driver Code
 
arr = [2, 4, 5, 5, 3, 5];
N = len(arr)
 
# Function Call
print(checkContiguous(arr, N));
 
# This code is contributed by Saurabh Jaiswal

C#

// C# code for the above approach
using System;
class GFG {
 
  // Function to check whether duplicate
  // elements in array arr[] are contiguous or not
  static String checkContiguous(int[] arr, int n)
  {
    int i;
 
    // Map to keep track of elements
    int[] visited = new int[n];
 
    for (i = 1; i < n; i++) {
      if (arr[i] == arr[i - 1])
        continue;
      else if (visited[arr[i]] == 0)
        return "No";
      visited[arr[i]] = 1;
    }
 
    return "Yes";
  }
 
  // Driver Code
  public static void Main()
  {
 
    int[] arr = { 2, 4, 5, 5, 3, 5 };
    int N = arr.Length;
 
    // Function Call
    Console.WriteLine(checkContiguous(arr, N));
  }
}
 
// This code is contributed by ukasp.

Javascript

输出

No

时间复杂度： O(N)
辅助空间： O(N)