C#程序查找出现的奇数次 _C#开发

本文概述

给定一个正整数数组。除一个数字为奇数次外, 所有数字均出现偶数次。在O(n)时间和恒定空间中找到数字。
例子：

Input : arr = {1, 2, 3, 2, 3, 1, 3} Output : 3Input : arr = {5, 7, 2, 7, 5, 2, 5} Output : 5

//C# program to find the element //occurring odd number of times using System; class GFG { //Function to find the element //occurring odd number of times static int getOddOccurrence( int [] arr, int arr_size) { for ( int i = 0; i < arr_size; i++) { int count = 0; for ( int j = 0; j < arr_size; j++) { if (arr[i] == arr[j]) count++; } if (count % 2 != 0) return arr[i]; } return -1; }//Driver code public static void Main() { int [] arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 }; int n = arr.Length; Console.Write(getOddOccurrence(arr, n)); } }//This code is contributed by Sam007

【C#程序查找出现的奇数次】输出如下：

5

一种更好的解决方案是使用哈希。使用数组元素作为键, 并将其计数作为值。创建一个空的哈希表。一对一遍历给定的数组元素并存储计数。该解决方案的时间复杂度为O(n)。但是它需要额外的空间来进行哈希处理。
程序：
C#

//C# program to find the element //occurring odd number of times using System; class GFG { //Function to find the element //occurring odd number of times static int getOddOccurrence( int [] arr, int arr_size) { int res = 0; for ( int i = 0; i < arr_size; i++) { res = res ^ arr[i]; } return res; }//Driver code public static void Main() { int [] arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 }; int n = arr.Length; Console.Write(getOddOccurrence(arr, n)); } }//This code is contributed by Sam007

输出如下：