给定一个未排序的数组和一个数字n，请查找数组中是否存在一对差为n的元素。
例子：
Input: arr[] = {5, 20, 3, 2, 50, 80}, n = 78
Output: Pair Found: (2, 80)
Input: arr[] = {90, 70, 20, 80, 50}, n = 45
Output: No Such Pair
最简单的方法是运行两个循环，外循环选择第一个元素(较小的元素)，内循环查找外循环加上n拾取的元素。该方法的时间复杂度为O(n ^ 2)。
我们可以使用排序和二进制搜索将时间复杂度提高到O(nLogn)。第一步是按升序对数组进行排序。数组排序后，从左到右遍历该数组，对于每个元素arr [i]，在arr [i + 1..n-1]中对arr [i] + n进行二进制搜索。如果找到该元素，则返回该对。
第一步和第二步都采用O(nLogn)。因此，总体复杂度为O(nLogn)。
上述算法的第二步可以提高到O(n)。第一步保持不变。第二步的想法是采用两个索引变量i和j，分别将它们初始化为0和1。现在运行一个线性循环。如果arr [j] – arr [i]小于n，我们需要寻找更大的arr [j]，因此增加j。如果arr [j] – arr [i]大于n，我们需要寻找更大的arr [i]，因此增加i。感谢Aashish Barnwal提出了这种方法。
以下代码仅用于算法的第二步，它假定数组已经排序。
C++
// C++ program to find a pair with the given difference
#include
using namespace std;
// The function assumes that the array is sorted
bool findPair(int arr[], int size, int n)
{
// Initialize positions of two elements
int i = 0;
int j = 1;
// Search for a pair
while (i < size && j < size)
{
if (i != j && arr[j] - arr[i] == n)
{
cout << "Pair Found: (" << arr[i] <<
", " << arr[j] << ")";
return true;
}
else if (arr[j]-arr[i] < n)
j++;
else
i++;
}
cout << "No such pair";
return false;
}
// Driver program to test above function
int main()
{
int arr[] = {1, 8, 30, 40, 100};
int size = sizeof(arr)/sizeof(arr[0]);
int n = 60;
findPair(arr, size, n);
return 0;
}
// This is code is contributed by rathbhupendra C
// C program to find a pair with the given difference
#include
// The function assumes that the array is sorted
bool findPair(int arr[], int size, int n)
{
// Initialize positions of two elements
int i = 0;
int j = 1;
// Search for a pair
while (i Java
// Java program to find a pair with the given difference
import java.io.*;
class PairDifference
{
// The function assumes that the array is sorted
static boolean findPair(int arr[],int n)
{
int size = arr.length;
// Initialize positions of two elements
int i = 0, j = 1;
// Search for a pair
while (i < size && j < size)
{
if (i != j && arr[j]-arr[i] == n)
{
System.out.print("Pair Found: "+
"( "+arr[i]+", "+ arr[j]+" )");
return true;
}
else if (arr[j] - arr[i] < n)
j++;
else
i++;
}
System.out.print("No such pair");
return false;
}
// Driver program to test above function
public static void main (String[] args)
{
int arr[] = {1, 8, 30, 40, 100};
int n = 60;
findPair(arr,n);
}
}
/*This code is contributed by Devesh Agrawal*/Python
# Python program to find a pair with the given difference
# The function assumes that the array is sorted
def findPair(arr,n):
size = len(arr)
# Initialize positions of two elements
i,j = 0,1
# Search for a pair
while i < size and j < size:
if i != j and arr[j]-arr[i] == n:
print "Pair found (",arr[i],",",arr[j],")"
return True
elif arr[j] - arr[i] < n:
j+=1
else:
i+=1
print "No pair found"
return False
# Driver function to test above function
arr = [1, 8, 30, 40, 100]
n = 60
findPair(arr, n)
# This code is contributed by Devesh AgrawalC#
// C# program to find a pair with the given difference
using System;
class GFG {
// The function assumes that the array is sorted
static bool findPair(int []arr, int n)
{
int size = arr.Length;
// Initialize positions of two elements
int i = 0, j = 1;
// Search for a pair
while (i < size && j < size)
{
if (i != j && arr[j] - arr[i] == n)
{
Console.Write("Pair Found: "
+ "( " + arr[i] + ", " + arr[j] +" )");
return true;
}
else if (arr[j] - arr[i] < n)
j++;
else
i++;
}
Console.Write("No such pair");
return false;
}
// Driver program to test above function
public static void Main ()
{
int []arr = {1, 8, 30, 40, 100};
int n = 60;
findPair(arr, n);
}
}
// This code is contributed by Sam007.PHP
输出：
Pair Found: (40, 100)散列也可以用来解决此问题。创建一个空的哈希表HT。遍历数组，将数组元素用作哈希键，然后在HT中输入它们。再次遍历数组，寻找HT中的值n + arr [i]。