# 連續元素排序數組中僅重復元素的計數 > 原文： [https://www.geeksforgeeks.org/count-of-only-repeated-element-in-a-sorted-array-of-consecutive-elements/](https://www.geeksforgeeks.org/count-of-only-repeated-element-in-a-sorted-array-of-consecutive-elements/) 給定一個連續元素的排序數組。 該數組只有一個元素重復多次。 任務是找到重復元素序列的長度。 預期時間復雜度：小于 0（n） **示例**： ``` 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 ``` 我們需要找到兩件事： 1. 元素重復的次數。 如果對數組排序，并且兩個相鄰元素的最大差為 1，則重復序列的長度為 n – 1 –（array [n-1] – array [0]） 2. 元素的值。 對于該值，我們執行[二分搜索](https://www.geeksforgeeks.org/binary-search/)。 ## C++ ```cpp // C++ program to find the only repeated element // and number of times it appears #include <bits/stdc++.h> using namespace std; // Assumptions : vector a is sorted, max-difference // of two adjacent elements is 1 pair<int, int> sequence(const vector<int>& 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() { ????pair<int, int>p = sequence({ 1, 2, 3, 4, 4, 4, 5, 6 }); ????cout << "Repeated element is " << p.first ?????????<< ", it appears " << p.second << " times"; ????return 0; } ``` ## Java ```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<Integer> 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 ```py # 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# ```cs // 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<int> 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<int>(array)); ????????Console.WriteLine("Repeated element is " + p.first + ????????????????????????", it appears " + p.second + " times"); ????} } // This code has been contributed by 29AjayKumar ``` **輸出**： ``` Repeated element is 4, it appears 3 times ```