连续元素排序数组中仅重复元素的计数

给定一个连续元素的排序数组。该数组只有一个元素重复多次。任务是找到重复元素序列的长度。

预期时间复杂度：小于0(n)

例子：

Input  : arr[] = {1, 2, 3, 4, 4, 4, 5, 6}
Output : 4 3
Repeated element is 4 and it appears 3 times.

Input  : arr[] = {4, 4, 4, 4, 4}
Output : 4 5 

Input  : arr[] = {6, 7, 8, 9, 10, 10, 11}
Output : 10 2

Input  : arr[] = {6, 7, 8, 9, 10, 10, 10}
Output : 10 3

我们需要找到两件事：

1. 元素重复的次数。如果对数组进行排序并且两个相邻元素的最大差为1，则重复序列的长度为n – 1 –(array [n-1] – array [0])
2. 元素的值。对于值，我们进行二进制搜索。

C++

// C++ program to find the only repeated element
// and number of times it appears
#include 
using namespace std;
  
// Assumptions : vector a is sorted, max-difference
// of two adjacent elements is 1
pair sequence(const vector& a)
{
    if (a.size() == 0)
        return {0, 0};
  
    int s = 0;
    int e = a.size() - 1;
    while (s < e)
    {
        int m = (s + e) / 2;
  
        // if a[m] = m + a[0], there is no
        // repeating character in [s..m]
        if (a[m] >= m + a[0])
            s = m + 1;
  
       // if a[m] < m + a[0], there is a
       // repeating character in [s..m]
        else
            e = m;
    }
    return {a[s], a.size() - (a[a.size() - 1] - a[0])};
}
  
// Driver code
int main()
{
    pairp = sequence({ 1, 2, 3, 4, 4, 4, 5, 6 });
    cout << "Repeated element is " << p.first
         << ", it appears " << p.second << " times";
    return 0;
}

Java

// Java program to find the only repeated element
// and number of times it appears
  
import java.awt.Point;
import java.util.Arrays;
import java.util.Vector;
  
class Test
{
    // Assumptions : vector a is sorted, max-difference
    // of two adjacent elements is 1
    static Point sequence(Vector a)
    {
        if (a.size() == 0)
            return new Point(0, 0);
       
        int s = 0;
        int e = a.size() - 1;
        while (s < e)
        {
            int m = (s + e) / 2;
       
            // if a[m] = m + a[0], there is no
            // repeating character in [s..m]
            if (a.get(m) >= m + a.get(0))
                s = m + 1;
       
           // if a[m] < m + a[0], there is a
           // repeating character in [s..m]
            else
                e = m;
        }
        return new Point(a.get(s), a.size() - (a.get(a.size() - 1) - a.get(0)));
    }
      
    // Driver method
    public static void main(String args[])
    {
        Integer array[] = new Integer[]{1, 2, 3, 4, 4, 4, 5, 6};
        Point p = sequence(new Vector<>(Arrays.asList(array)));
        System.out.println("Repeated element is " + p.x +
                           ", it appears " + p.y + " times");
    }
}

Python3

# Python3 program to find the 
# only repeated element and 
# number of times it appears
  
# Assumptions : vector a is sorted, 
# max-difference of two adjacent
# elements is 1
def sequence(a):
    if (len(a) == 0):
        return [0, 0]
  
    s = 0
    e = len(a) - 1
    while (s < e):
        m = (s + e) // 2
  
        # if a[m] = m + a[0], there is no
        # repeating character in [s..m]
        if (a[m] >= m + a[0]):
            s = m + 1
  
        # if a[m] < m + a[0], there is a
        # repeating character in [s..m]
        else:
            e = m
    return [a[s], len(a) - (
                a[len(a) - 1] - a[0])]
  
# Driver code
p = sequence([1, 2, 3, 4, 4, 4, 5, 6])
print("Repeated element is", p[0], 
             ", it appears", p[1], "times")
  
# This code is contributed by Mohit Kumar

C#

// C# program to find the only repeated element
// and number of times it appears
using System;
using System.Collections.Generic;
  
public class Point 
{ 
    public int first, second; 
  
        public Point(int first, int second) 
        { 
            this.first = first; 
            this.second = second; 
        } 
      
      
} 
public class Test
{
    // Assumptions : vector a is sorted, max-difference
    // of two adjacent elements is 1
    static Point sequence(List a)
    {
        if (a.Count == 0)
            return new Point(0, 0);
      
        int s = 0;
        int e = a.Count - 1;
        while (s < e)
        {
            int m = (s + e) / 2;
      
            // if a[m] = m + a[0], there is no
            // repeating character in [s..m]
            if (a[m] >= m + a[0])
                s = m + 1;
      
        // if a[m] < m + a[0], there is a
        // repeating character in [s..m]
            else
                e = m;
        }
        return new Point(a[s], a.Count - (a[a.Count - 1] - a[0]));
    }
      
    // Driver code
    public static void Main(String []args)
    {
        int []array = new int[]{1, 2, 3, 4, 4, 4, 5, 6};
        Point p = sequence(new List(array));
        Console.WriteLine("Repeated element is " + p.first +
                        ", it appears " + p.second + " times");
    }
}
  
// This code has been contributed by 29AjayKumar

输出：

Repeated element is 4, it appears 3 times