打印排序数组中存在的所有唯一元素

给定一个大小为N的排序数组arr[] ，任务是打印数组中的所有唯一元素。

An array element is said to be unique if the frequency of that element in the array is 1.

例子：

Input: arr[ ] = {1, 1, 2, 2, 3, 4, 5, 5}
Output: 3 4
Explanation: Since 1, 2, 5 are occurring more than once in the array, the distinct elements are 3 and 4.

Input: arr[ ] = {1, 2, 3, 3, 3, 4, 5, 6}
Output: 1 2 4 5 6

方法：解决问题的最简单方法是遍历数组arr[]并仅打印频率为1的那些元素。请按照以下步骤解决问题：

• 遍历数组arr[]并初始化一个变量，比如cnt = 0，以计算当前数组元素的频率。
• 由于数组已经排序，请检查当前元素是否与前一个元素相同。如果发现为真，则更新cnt += 1 。
• 否则，如果cnt = 1，则打印元素。否则，继续。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include 
using namespace std;
 
// Function to print all unique
// elements present in a sorted array
void RemoveDuplicates(int arr[], int n)
{
 
    int i = 0;
 
    // Traverse the array
    while (i < n) {
 
        int cur = arr[i];
 
        // Stores frequency of
        // the current element
        int cnt = 0;
 
        // Iterate until end of the
        // array is reached or current
        // element is not the same as the
        // previous element
        while (i < n and cur == arr[i]) {
            cnt++;
            i++;
        }
 
        // If current element is unique
        if (cnt == 1) {
 
            cout << cur << " ";
        }
    }
}
 
// Driver Code
int main()
{
 
    // Given Input
    int arr[] = { 1, 3, 3, 5, 5, 6, 10 };
    int N = 7;
 
    // Function Call
    RemoveDuplicates(arr, N);
 
    return 0;
}

Java

// Java Program for the above approach
import java.io.*;
 
class GFG
{
   
  // Function to print all unique
  // elements present in a sorted array
  static void RemoveDuplicates(int arr[], int n)
  {
 
    int i = 0;
 
    // Traverse the array
    while (i < n) {
 
      int cur = arr[i];
 
      // Stores frequency of
      // the current element
      int cnt = 0;
 
      // Iterate until end of the
      // array is reached or current
      // element is not the same as the
      // previous element
      while (i < n && cur == arr[i]) {
        cnt++;
        i++;
      }
 
      // If current element is unique
      if (cnt == 1) {
        System.out.print(cur +" ");
      }
    }
  }
 
  // Driver Code
 
  public static void main (String[] args)
  {
 
    // Given Input
    int arr[] = { 1, 3, 3, 5, 5, 6, 10 };
    int N = 7;
 
    // Function Call
    RemoveDuplicates(arr, N);
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Function to print all unique
# elements present in a sorted array
def RemoveDuplicates(arr, n):
    i = 0
    while i < n:
        cur = arr[i]
         
        # Stores frequency of
        # the current element
        cnt = 0
 
        # Iterate until end of the
        # array is reached or current
        # element is not the same as the
        # previous element
        while i < n and cur == arr[i]:
            cnt += 1
            i += 1
        if cnt == 1:
            print(cur, end=" ")
 
# Driver code
if __name__ == "__main__":
   
    # Given Input
    arr = [1, 3, 3, 5, 5, 6, 10]
    N = 7
 
    # Function Call
    RemoveDuplicates(arr, N)
     
# This code is contributed by Kushagra Bansal

C#

// C# Program for the above approach
using System;
 
class GFG {
 
    // Function to print all unique
    // elements present in a sorted array
    static void RemoveDuplicates(int[] arr, int n)
    {
 
        int i = 0;
 
        // Traverse the array
        while (i < n) {
 
            int cur = arr[i];
 
            // Stores frequency of
            // the current element
            int cnt = 0;
 
            // Iterate until end of the
            // array is reached or current
            // element is not the same as the
            // previous element
            while (i < n && cur == arr[i]) {
                cnt++;
                i++;
            }
 
            // If current element is unique
            if (cnt == 1) {
                Console.Write(cur + " ");
            }
        }
    }
 
    // Driver Code
 
    public static void Main()
    {
 
        // Given Input
        int[] arr = { 1, 3, 3, 5, 5, 6, 10 };
        int N = 7;
 
        // Function Call
        RemoveDuplicates(arr, N);
    }
}
 
// This code is contributed by rishavmahato348.

Javascript

       // JavaScript Program for the above approach
 
       // Function to print all unique
       // elements present in a sorted array
       function RemoveDuplicates(arr, n) {
 
           let i = 0;
 
           // Traverse the array
           while (i < n) {
 
               let cur = arr[i];
 
               // Stores frequency of
               // the current element
               let cnt = 0;
 
               // Iterate until end of the
               // array is reached or current
               // element is not the same as the
               // previous element
               while (i < n && cur == arr[i]) {
                   cnt++;
                   i++;
               }
 
               // If current element is unique
               if (cnt == 1) {
 
                   document.write(cur + " ");
               }
           }
       }
 
       // Driver Code
 
       // Given Input
       let arr = [1, 3, 3, 5, 5, 6, 10];
       let N = 7;
 
       // Function Call
       RemoveDuplicates(arr, N);
 
   // This code is contributed by Potta Lokesh
 
   

输出

1 6 10 

时间复杂度： O(N)
辅助空间： O(1)