给定一个排序的数组,其中所有元素出现两次(一个接一个),而一个元素仅出现一次。在O(log n)复杂度中找到该元素。
例子:
Input: arr[] = {1, 1, 3, 3, 4, 5, 5, 7, 7, 8, 8}
Output: 4
Input: arr[] = {1, 1, 3, 3, 4, 4, 5, 5, 7, 7, 8}
Output: 8
一个简单的解决方案是从左到右遍历数组。由于数组已排序,因此我们可以轻松找出所需的元素。
下面是上述方法的实现。
C++
// C++ program to find the element that
// appears only once
#include
using namespace std;
// A Linear Search based function to find
// the element that appears only once
void search(int arr[], int n)
{
int ans = -1;
for (int i = 0; i < n; i += 2) {
if (arr[i] != arr[i + 1]) {
ans = arr[i];
break;
}
}
if (arr[n - 2] != arr[n - 1])
ans = arr[n-1];
// ans = -1 if no such element is present.
cout << "The required element is " << ans << "\n";
}
// Driver code
int main()
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = sizeof(arr) / sizeof(arr[0]);
search(arr, len);
return 0;
}
// This code is contributed by yashbeersingh42
Java
// Java program to find the element that
// appears only once
import java.io.*;
class GFG {
// A Linear Search based function to find
// the element that appears only once
static void search(int arr[], int n)
{
int ans = -1;
for (int i = 0; i < n; i += 2) {
if (arr[i] != arr[i + 1]) {
ans = arr[i];
break;
}
}
if (arr[n - 2] != arr[n - 1])
ans = arr[n-1];
// ans = -1 if no such element is present.
System.out.println("The required element is "
+ ans);
}
public static void main(String[] args)
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = arr.length;
search(arr, len);
}
}
Python3
# Python3 program to find the element that
# appears only once
# A Linear Search based function to find
# the element that appears only once
def search(arr, n):
ans = -1
for i in range(0, n, 2):
if (arr[i] != arr[i + 1]):
ans = arr[i]
break
if(arr[n-2] != arr[n-1]):
ans = arr[n-1]
# ans = -1 if no such element is present.
print("The required element is", ans)
# Driver code
arr = [1, 1, 2, 4, 4, 5, 5, 6, 6]
Len = len(arr)
search(arr, Len)
# This code is contributed by divyesh072019
C#
// C# program to find the element that
// appears only once
using System;
class GFG {
// A Linear Search based function to find
// the element that appears only once
static void search(int[] arr, int n)
{
int ans = -1;
for (int i = 0; i < n; i += 2) {
if (arr[i] != arr[i + 1]) {
ans = arr[i];
break;
}
}
if (arr[n - 2] != arr[n - 1])
ans = arr[n-1];
// ans = -1 if no such element is present.
Console.Write("The required element is "
+ ans);
}
public static void Main(String[] args)
{
int[] arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = arr.Length;
search(arr, len);
}
}
// This code is contributed by shivanisinghss2110
Javascript
C++
// C++ program to find the element that
// appears only once
#include
using namespace std;
// A XOR based function to find
// the element that appears only once
void search(int arr[], int n)
{
int XOR = 0;
for (int i = 0; i < n; i++) {
XOR = XOR ^ arr[i];
}
cout << "The required element is " << XOR << "\n";
}
// Driver code
int main()
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = sizeof(arr) / sizeof(arr[0]);
search(arr, len);
return 0;
}
// This code is contributed by yashbeersingh42
Java
// Java program to find the element that
// appears only once
import java.io.*;
class GFG {
// A XOR based function to find
// the element that appears only once
static void search(int arr[], int n)
{
int XOR = 0;
for (int i = 0; i < n; i++) {
XOR = XOR ^ arr[i];
}
System.out.println("The required element is "
+ XOR);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = arr.length;
search(arr, len);
}
}
// This code is contributed by yashbeersingh42
Python3
# Python3 program to find the element that
# appears only once
# A XOR based function to find
# the element that appears only once
def search(arr, n) :
XOR = 0
for i in range(n) :
XOR = XOR ^ arr[i]
print("The required element is", XOR)
# Driver code
arr = [ 1, 1, 2, 4, 4, 5, 5, 6, 6 ]
Len = len(arr)
search(arr, Len)
# This code is contributed by divyesh072019
C#
// C# program to find the element that
// appears only once
using System;
class GFG{
// A XOR based function to find
// the element that appears only once
static void search(int []arr, int n)
{
int XOR = 0;
for(int i = 0; i < n; i++)
{
XOR = XOR ^ arr[i];
}
Console.Write("The required element is " + XOR);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = arr.Length;
search(arr, len);
}
}
// This code is contributed by shivanisinghss2110
C++
// C++ program to find the element that
// appears only once
#include
using namespace std;
// A Binary Search based function to find
// the element that appears only once
void search(int arr[], int low, int high)
{
// Base cases
if (low > high)
return;
if (low == high) {
cout << "The required element is " << arr[low];
return;
}
// Find the middle point
int mid = (low + high) / 2;
// If mid is even and element next to mid is
// same as mid, then output element lies on
// right side, else on left side
if (mid % 2 == 0) {
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
// If mid is odd
else {
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
// Driver code
int main()
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = sizeof(arr) / sizeof(arr[0]);
search(arr, 0, len - 1);
return 0;
}
// This code is contributed by ShubhamCoder
C
// C program to find the element that appears only once
#include
// A Binary Search based function to find the element
// that appears only once
void search(int* arr, int low, int high)
{
// Base cases
if (low > high)
return;
if (low == high) {
printf("The required element is %d ", arr[low]);
return;
}
// Find the middle point
int mid = (low + high) / 2;
// If mid is even and element next to mid is
// same as mid, then output element lies on
// right side, else on left side
if (mid % 2 == 0) {
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
else // If mid is odd
{
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
// Driver code
int main()
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = sizeof(arr) / sizeof(arr[0]);
search(arr, 0, len - 1);
return 0;
}
Java
// Java program to find the element that appears only once
public class Main {
// A Binary Search based method to find the element
// that appears only once
public static void search(int[] arr, int low, int high)
{
if (low > high)
return;
if (low == high) {
System.out.println("The required element is "
+ arr[low]);
return;
}
// Find the middle point
int mid = (low + high) / 2;
// If mid is even and element next to mid is
// same as mid, then output element lies on
// right side, else on left side
if (mid % 2 == 0) {
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
// If mid is odd
else if (mid % 2 == 1) {
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
search(arr, 0, arr.length - 1);
}
}
// This code is contributed by Tanisha Mittal
Python
# A Binary search based function to find
# the element that appears only once
def search(arr, low, high):
# Base cases
if low > high:
return None
if low == high:
return arr[low]
# Find the middle point
mid = low + (high - low)/2
# If mid is even and element next to mid is
# same as mid, then output element lies on
# right side, else on left side
if mid % 2 == 0:
if arr[mid] == arr[mid+1]:
return search(arr, mid+2, high)
else:
return search(arr, low, mid)
else:
# if mid is odd
if arr[mid] == arr[mid-1]:
return search(arr, mid+1, high)
else:
return search(arr, low, mid-1)
# Driver Code
# Test Array
arr = [1, 1, 2, 4, 4, 5, 5, 6, 6]
# Function call
result = search(arr, 0, len(arr)-1)
if result is not None:
print "The required element is %d" % result
else:
print "Invalid Array"
C#
// C# program to find the element
// that appears only once
using System;
class GFG {
// A Binary Search based
// method to find the element
// that appears only once
public static void search(int[] arr, int low, int high)
{
if (low > high)
return;
if (low == high) {
Console.WriteLine("The required element is "
+ arr[low]);
return;
}
// Find the middle point
int mid = (low + high) / 2;
// If mid is even and element
// next to mid is same as mid
// then output element lies on
// right side, else on left side
if (mid % 2 == 0) {
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
// If mid is odd
else if (mid % 2 == 1) {
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
search(arr, 0, arr.Length - 1);
}
}
// This code is contributed by Nitin Mittal.
PHP
$high)
return;
if ($low==$high)
{
echo("The required element is " );
echo $arr[$low] ;
return;
}
// Find the middle point
$mid = ($low + $high) / 2;
// If mid is even and element
// next to mid is same as mid,
// then output element lies on
// right side, else on left side
if ($mid % 2 == 0)
{
if ($arr[$mid] == $arr[$mid + 1])
search($arr, $mid + 2, $high);
else
search($arr, $low, $mid);
}
// If mid is odd
else
{
if ($arr[$mid] == $arr[$mid - 1])
search($arr, $mid + 1, $high);
else
search($arr, $low, $mid - 1);
}
}
// Driver Code
$arr = array(1, 1, 2, 4, 4, 5, 5, 6, 6);
$len = sizeof($arr);
search($arr, 0, $len - 1);
// This code is contributed by nitin mittal
?>
输出
The required element is 2
时间复杂度: O(n)
空间复杂度:O(1)
另一个简单的解决方案是使用XOR的属性(a ^ a = 0&a ^ 0 = a)。查找完整数组的XOR的想法。数组的XOR是必需的答案。
下面是上述方法的实现。
C++
// C++ program to find the element that
// appears only once
#include
using namespace std;
// A XOR based function to find
// the element that appears only once
void search(int arr[], int n)
{
int XOR = 0;
for (int i = 0; i < n; i++) {
XOR = XOR ^ arr[i];
}
cout << "The required element is " << XOR << "\n";
}
// Driver code
int main()
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = sizeof(arr) / sizeof(arr[0]);
search(arr, len);
return 0;
}
// This code is contributed by yashbeersingh42
Java
// Java program to find the element that
// appears only once
import java.io.*;
class GFG {
// A XOR based function to find
// the element that appears only once
static void search(int arr[], int n)
{
int XOR = 0;
for (int i = 0; i < n; i++) {
XOR = XOR ^ arr[i];
}
System.out.println("The required element is "
+ XOR);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = arr.length;
search(arr, len);
}
}
// This code is contributed by yashbeersingh42
Python3
# Python3 program to find the element that
# appears only once
# A XOR based function to find
# the element that appears only once
def search(arr, n) :
XOR = 0
for i in range(n) :
XOR = XOR ^ arr[i]
print("The required element is", XOR)
# Driver code
arr = [ 1, 1, 2, 4, 4, 5, 5, 6, 6 ]
Len = len(arr)
search(arr, Len)
# This code is contributed by divyesh072019
C#
// C# program to find the element that
// appears only once
using System;
class GFG{
// A XOR based function to find
// the element that appears only once
static void search(int []arr, int n)
{
int XOR = 0;
for(int i = 0; i < n; i++)
{
XOR = XOR ^ arr[i];
}
Console.Write("The required element is " + XOR);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = arr.Length;
search(arr, len);
}
}
// This code is contributed by shivanisinghss2110
输出
The required element is 2
时间复杂度: O(n)
空间复杂度: O(1)
一个有效的解决方案可以在O(Log n)时间中找到所需的元素。这个想法是使用二进制搜索。下面是对输入数组的观察。
要求之前的所有元素的第一个出现在偶数索引(0,2,..),下一个出现在奇数索引(1,3,…)。并且在必需元素之后的所有元素在奇数索引处第一个出现,在偶数索引处下一个出现。
1)找到中间索引,说’mid’。
2)如果“ mid”是偶数,则比较arr [mid]和arr [mid + 1]。如果两者相同,则必需的元素在“ mid”之后,否则在mid之前。
3)如果’mid’为奇数,则比较arr [mid]和arr [mid – 1]。如果两者相同,则必需的元素在“ mid”之后,否则在mid之前。
下面是基于以上思想的实现:
C++
// C++ program to find the element that
// appears only once
#include
using namespace std;
// A Binary Search based function to find
// the element that appears only once
void search(int arr[], int low, int high)
{
// Base cases
if (low > high)
return;
if (low == high) {
cout << "The required element is " << arr[low];
return;
}
// Find the middle point
int mid = (low + high) / 2;
// If mid is even and element next to mid is
// same as mid, then output element lies on
// right side, else on left side
if (mid % 2 == 0) {
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
// If mid is odd
else {
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
// Driver code
int main()
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = sizeof(arr) / sizeof(arr[0]);
search(arr, 0, len - 1);
return 0;
}
// This code is contributed by ShubhamCoder
C
// C program to find the element that appears only once
#include
// A Binary Search based function to find the element
// that appears only once
void search(int* arr, int low, int high)
{
// Base cases
if (low > high)
return;
if (low == high) {
printf("The required element is %d ", arr[low]);
return;
}
// Find the middle point
int mid = (low + high) / 2;
// If mid is even and element next to mid is
// same as mid, then output element lies on
// right side, else on left side
if (mid % 2 == 0) {
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
else // If mid is odd
{
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
// Driver code
int main()
{
int arr[] = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
int len = sizeof(arr) / sizeof(arr[0]);
search(arr, 0, len - 1);
return 0;
}
Java
// Java program to find the element that appears only once
public class Main {
// A Binary Search based method to find the element
// that appears only once
public static void search(int[] arr, int low, int high)
{
if (low > high)
return;
if (low == high) {
System.out.println("The required element is "
+ arr[low]);
return;
}
// Find the middle point
int mid = (low + high) / 2;
// If mid is even and element next to mid is
// same as mid, then output element lies on
// right side, else on left side
if (mid % 2 == 0) {
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
// If mid is odd
else if (mid % 2 == 1) {
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
search(arr, 0, arr.length - 1);
}
}
// This code is contributed by Tanisha Mittal
Python
# A Binary search based function to find
# the element that appears only once
def search(arr, low, high):
# Base cases
if low > high:
return None
if low == high:
return arr[low]
# Find the middle point
mid = low + (high - low)/2
# If mid is even and element next to mid is
# same as mid, then output element lies on
# right side, else on left side
if mid % 2 == 0:
if arr[mid] == arr[mid+1]:
return search(arr, mid+2, high)
else:
return search(arr, low, mid)
else:
# if mid is odd
if arr[mid] == arr[mid-1]:
return search(arr, mid+1, high)
else:
return search(arr, low, mid-1)
# Driver Code
# Test Array
arr = [1, 1, 2, 4, 4, 5, 5, 6, 6]
# Function call
result = search(arr, 0, len(arr)-1)
if result is not None:
print "The required element is %d" % result
else:
print "Invalid Array"
C#
// C# program to find the element
// that appears only once
using System;
class GFG {
// A Binary Search based
// method to find the element
// that appears only once
public static void search(int[] arr, int low, int high)
{
if (low > high)
return;
if (low == high) {
Console.WriteLine("The required element is "
+ arr[low]);
return;
}
// Find the middle point
int mid = (low + high) / 2;
// If mid is even and element
// next to mid is same as mid
// then output element lies on
// right side, else on left side
if (mid % 2 == 0) {
if (arr[mid] == arr[mid + 1])
search(arr, mid + 2, high);
else
search(arr, low, mid);
}
// If mid is odd
else if (mid % 2 == 1) {
if (arr[mid] == arr[mid - 1])
search(arr, mid + 1, high);
else
search(arr, low, mid - 1);
}
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 1, 1, 2, 4, 4, 5, 5, 6, 6 };
search(arr, 0, arr.Length - 1);
}
}
// This code is contributed by Nitin Mittal.
的PHP
$high)
return;
if ($low==$high)
{
echo("The required element is " );
echo $arr[$low] ;
return;
}
// Find the middle point
$mid = ($low + $high) / 2;
// If mid is even and element
// next to mid is same as mid,
// then output element lies on
// right side, else on left side
if ($mid % 2 == 0)
{
if ($arr[$mid] == $arr[$mid + 1])
search($arr, $mid + 2, $high);
else
search($arr, $low, $mid);
}
// If mid is odd
else
{
if ($arr[$mid] == $arr[$mid - 1])
search($arr, $mid + 1, $high);
else
search($arr, $low, $mid - 1);
}
}
// Driver Code
$arr = array(1, 1, 2, 4, 4, 5, 5, 6, 6);
$len = sizeof($arr);
search($arr, 0, $len - 1);
// This code is contributed by nitin mittal
?>
输出
The required element is 2
时间复杂度: O(Log n)
注意:这个问题其他的解决方案的办法在这篇文章中讨论的细微变化。