根据参考位置重新分配元素

考虑一个问题，其中可能一次又一次地搜索相同的元素。有效地执行搜索操作。

例子 ：

Input : arr[] = {12 25 36 85 98 75 89 15 63 66
                               64 74 27 83 97}
          q[] = {63, 63, 86, 63, 78}
Output : Yes Yes No Yes No
We need one by one search items of q[] in arr[].
The element 63 is present, 78 and 86 are not present.

这个想法很简单，我们将搜索到的元素移到数组的前面，以便下次可以快速搜索它。

C++

// C++ program to implement search for an item
// that is searched again and again.
#include 
using namespace std;
  
// A function to perform sequential search.
bool search(int arr[], int n, int x)
{
    // Linearly search the element
    int res = -1;
    for (int i = 0; i < n; i++) 
        if (x == arr[i])
        res = i; 
  
    // If not found
    if (res == -1)
        return false; 
  
    // Shift elements before one position
    int temp = arr[res];
    for (int i = res; i > 0; i--)
        arr[i] = arr[i - 1];
  
    arr[0] = temp;
    return true;
}
  
// Driver Code
int main()
{
    int arr[] = { 12, 25, 36, 85, 98, 75, 89, 15,
                    63, 66, 64, 74, 27, 83, 97 };
    int q[] = {63, 63, 86, 63, 78};
    int n = sizeof(arr)/sizeof(arr[0]);
    int m = sizeof(q)/sizeof(q[0]);
    for (int i=0; i

---

Java

// Java program to implement search for an item
// that is searched again and again.
import java.util.*;
  
class solution
{
  
// A function to perform sequential search.
static boolean search(int[] arr, int n, int x)
{
    // Linearly search the element
    int res = -1;
    for (int i = 0; i < n; i++) 
        if (x == arr[i])
        res = i; 
  
    // If not found
    if (res == -1)
        return false; 
  
    // Shift elements before one position
    int temp = arr[res];
    for (int i = res; i > 0; i--)
        arr[i] = arr[i - 1];
  
    arr[0] = temp;
    return true;
}
  
// Driver Code
public static void main(String args[])
{
    int []arr = { 12, 25, 36, 85, 98, 75, 89, 15,
                    63, 66, 64, 74, 27, 83, 97 };
    int []q = {63, 63, 86, 63, 78};
    int n = arr.length;
    int m = q.length;
    for (int i=0; i

---

Python3

# Python 3 program to implement search for 
# an item that is searched again and again.
  
# A function to perform sequential search.
def search(arr, n, x):
      
    # Linearly search the element
    res = -1
    for i in range(0, n, 1):
        if (x == arr[i]):
            res = i 
  
    # If not found
    if (res == -1):
        return False
  
    # Shift elements before 
    # one position
    temp = arr[res]
    i = res
    while(i > 0):
        arr[i] = arr[i - 1]
        i -= 1
  
    arr[0] = temp
    return True
  
# Driver Code
if __name__ == '__main__':
    arr = [12, 25, 36, 85, 98, 75, 89, 
           15, 63, 66, 64, 74, 27, 83, 97]
    q = [63, 63, 86, 63, 78]
    n = len(arr)
    m = len(q)
    for i in range(0, m, 1):
        if(search(arr, n, q[i])):
            print("Yes", end = " ")
        else:
            print("No", end = " ")
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to implement search for an 
// item that is searched again and again.
using System;
  
class GFG
{
      
// A function to perform sequential search.
static bool search(int[] arr, int n, int x)
{
    // Linearly search the element
    int res = -1;
    for (int i = 0; i < n; i++) 
        if (x == arr[i])
        res = i; 
  
    // If not found
    if (res == -1)
        return false; 
  
    // Shift elements before one position
    int temp = arr[res];
    for (int i = res; i > 0; i--)
        arr[i] = arr[i - 1];
  
    arr[0] = temp;
    return true;
}
  
// Driver Code
public static void Main()
{
    int[] arr = { 12, 25, 36, 85, 98, 75, 89, 15,
                      63, 66, 64, 74, 27, 83, 97 };
    int[] q = {63, 63, 86, 63, 78};
    int n = arr.Length;
    int m = q.Length;
    for (int i = 0; i < m; i++)
    {
        if(search(arr, n, q[i]) == true)
            Console.Write("Yes ");
        else
            Console.Write("No ");
    }
}
}
  
// This code is contributed by
// Akanksha Rai

PHP

 0; $i--)
        $arr[$i] = $arr[$i - 1];
  
    $arr[0] = $temp;
    return true;
}
  
// Driver Code
$arr = array(12, 25, 36, 85, 98, 75, 89, 15,
                 63, 66, 64, 74, 27, 83, 97);
$q = array(63, 63, 86, 63, 78);
$n = sizeof($arr);
$m = sizeof($q);
  
for ($i = 0; $i < $m; $i++)
    if(search($arr, $n, $q[$i]))
        echo "Yes ";
    else
        echo "No ";     
  
// This code is contributed
// by Akanksha Rai

输出：

Yes Yes No Yes No 

进一步考虑：通过使用链接列表，我们可以做得更好。在链接列表中，将项目移到最前面可以在O(1)时间内完成。
最好的解决方案是使用Splay Tree(为此目的而设计的数据结构)。 Splay树平均支持O(Log n)时间的插入，搜索和删除操作。此外，展开树是BST，因此我们可以按排序顺序快速打印元素。