插值搜索
给定 n 个均匀分布的值 arr[]，编写一个函数来搜索数组中的特定元素 x。 线性搜索在 O(n)时间内找到元素，跳转搜索用 O(√ n)时间，二分搜索法用 O(Log n)时间。 对于排序数组中的值均匀分布的情况，插值搜索是对二分搜索法的改进。二分搜索法总是去中间元素检查。另一方面，根据被搜索的关键字的值，插值搜索可以去往不同的位置。例如，如果关键字的值更接近最后一个元素，插值搜索可能会开始向末端搜索。 要找到要搜索的位置，它使用以下公式。
// The idea of formula is to return higher value of pos
// when element to be searched is closer to arr[hi]. And
// smaller value when closer to arr[lo]
pos = lo + [ (x-arr[lo])*(hi-lo) / (arr[hi]-arr[Lo]) ]
arr[] ==> Array where elements need to be searched
x ==> Element to be searched
lo ==> Starting index in arr[]
hi ==> Ending index in arr[]
pos 的公式可以推导如下。
Let's assume that the elements of the array are linearly distributed.
General equation of line : y = m*x + c.
y is the value in the array and x is its index.
Now putting value of lo,hi and x in the equation
arr[hi] = m*hi+c ----(1)
arr[lo] = m*lo+c ----(2)
x = m*pos + c ----(3)
m = (arr[hi] - arr[lo] )/ (hi - lo)
subtracting eqxn (2) from (3)
x - arr[lo] = m * (pos - lo)
lo + (x - arr[lo])/m = pos
pos = lo + (x - arr[lo]) *(hi - lo)/(arr[hi] - arr[lo])
算法 除了上面的分区逻辑，其余的插值算法都是一样的。 步骤 1: 在一个循环中，使用探针位置公式计算“pos”的值。 第二步:如果匹配，返回该项目的索引，退出。 步骤 3: 如果项目小于 arr[pos]，计算左子阵列的探头位置。否则，在右子数组中计算相同的值。 步骤 4: 重复，直到找到匹配项或子阵列减少到零。 下面是算法的实现。
C++
// C++ program to implement interpolation search
#include<bits/stdc++.h>
using namespace std;
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
int interpolationSearch(int arr[], int n, int x)
{
// Find indexes of two corners
int lo = 0, hi = (n - 1);
// Since array is sorted, an element present
// in array must be in range defined by corner
while (lo <= hi && x >= arr[lo] && x <= arr[hi])
{
if (lo == hi)
{
if (arr[lo] == x) return lo;
return -1;
}
// Probing the position with keeping
// uniform distribution in mind.
int pos = lo + (((double)(hi - lo) /
(arr[hi] - arr[lo])) * (x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in upper part
if (arr[pos] < x)
lo = pos + 1;
// If x is smaller, x is in the lower part
else
hi = pos - 1;
}
return -1;
}
// Driver Code
int main()
{
// Array of items on which search will
// be conducted.
int arr[] = {10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 18; // Element to be searched
int index = interpolationSearch(arr, n, x);
// If element was found
if (index != -1)
cout << "Element found at index " << index;
else
cout << "Element not found.";
return 0;
}
// This code is contributed by Mukul Singh.
C++
// C++ program to implement interpolation
// search with recursion
#include <bits/stdc++.h>
using namespace std;
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
int interpolationSearch(int arr[], int lo, int hi, int x)
{
int pos;
// Since array is sorted, an element present
// in array must be in range defined by corner
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo
+ (((double)(hi - lo) / (arr[hi] - arr[lo]))
* (x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if (arr[pos] < x)
return interpolationSearch(arr, pos + 1, hi, x);
// If x is smaller, x is in left sub array
if (arr[pos] > x)
return interpolationSearch(arr, lo, pos - 1, x);
}
return -1;
}
// Driver Code
int main()
{
// Array of items on which search will
// be conducted.
int arr[] = { 10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47 };
int n = sizeof(arr) / sizeof(arr[0]);
// Element to be searched
int x = 18;
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
cout << "Element found at index " << index;
else
cout << "Element not found.";
return 0;
}
// This code is contributed by equbalzeeshan
C
// C program to implement interpolation search
// with recursion
#include <stdio.h>
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
int interpolationSearch(int arr[], int lo, int hi, int x)
{
int pos;
// Since array is sorted, an element present
// in array must be in range defined by corner
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo
+ (((double)(hi - lo) / (arr[hi] - arr[lo]))
* (x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if (arr[pos] < x)
return interpolationSearch(arr, pos + 1, hi, x);
// If x is smaller, x is in left sub array
if (arr[pos] > x)
return interpolationSearch(arr, lo, pos - 1, x);
}
return -1;
}
// Driver Code
int main()
{
// Array of items on which search will
// be conducted.
int arr[] = { 10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47 };
int n = sizeof(arr) / sizeof(arr[0]);
int x = 18; // Element to be searched
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
printf("Element found at index %d", index);
else
printf("Element not found.");
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement interpolation
// search with recursion
import java.util.*;
class GFG {
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
public static int interpolationSearch(int arr[], int lo,
int hi, int x)
{
int pos;
// Since array is sorted, an element
// present in array must be in range
// defined by corner
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo
+ (((hi - lo) / (arr[hi] - arr[lo]))
* (x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if (arr[pos] < x)
return interpolationSearch(arr, pos + 1, hi,
x);
// If x is smaller, x is in left sub array
if (arr[pos] > x)
return interpolationSearch(arr, lo, pos - 1,
x);
}
return -1;
}
// Driver Code
public static void main(String[] args)
{
// Array of items on which search will
// be conducted.
int arr[] = { 10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47 };
int n = arr.length;
// Element to be searched
int x = 18;
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
System.out.println("Element found at index "
+ index);
else
System.out.println("Element not found.");
}
}
// This code is contributed by equbalzeeshan
计算机编程语言
# Python3 program to implement
# interpolation search
# with recursion
# If x is present in arr[0..n-1], then
# returns index of it, else returns -1.
def interpolationSearch(arr, lo, hi, x):
# Since array is sorted, an element present
# in array must be in range defined by corner
if (lo <= hi and x >= arr[lo] and x <= arr[hi]):
# Probing the position with keeping
# uniform distribution in mind.
pos = lo + ((hi - lo) // (arr[hi] - arr[lo]) *
(x - arr[lo]))
# Condition of target found
if arr[pos] == x:
return pos
# If x is larger, x is in right subarray
if arr[pos] < x:
return interpolationSearch(arr, pos + 1,
hi, x)
# If x is smaller, x is in left subarray
if arr[pos] > x:
return interpolationSearch(arr, lo,
pos - 1, x)
return -1
# Driver code
# Array of items in which
# search will be conducted
arr = [10, 12, 13, 16, 18, 19, 20,
21, 22, 23, 24, 33, 35, 42, 47]
n = len(arr)
# Element to be searched
x = 18
index = interpolationSearch(arr, 0, n - 1, x)
if index != -1:
print("Element found at index", index)
else:
print("Element not found")
# This code is contributed by Hardik Jain
C
// C# program to implement
// interpolation search
using System;
class GFG{
// If x is present in
// arr[0..n-1], then
// returns index of it,
// else returns -1.
static int interpolationSearch(int []arr, int lo,
int hi, int x)
{
int pos;
// Since array is sorted, an element
// present in array must be in range
// defined by corner
if (lo <= hi && x >= arr[lo] &&
x <= arr[hi])
{
// Probing the position
// with keeping uniform
// distribution in mind.
pos = lo + (((hi - lo) /
(arr[hi] - arr[lo])) *
(x - arr[lo]));
// Condition of
// target found
if(arr[pos] == x)
return pos;
// If x is larger, x is in right sub array
if(arr[pos] < x)
return interpolationSearch(arr, pos + 1,
hi, x);
// If x is smaller, x is in left sub array
if(arr[pos] > x)
return interpolationSearch(arr, lo,
pos - 1, x);
}
return -1;
}
// Driver Code
public static void Main()
{
// Array of items on which search will
// be conducted.
int []arr = new int[]{ 10, 12, 13, 16, 18,
19, 20, 21, 22, 23,
24, 33, 35, 42, 47 };
// Element to be searched
int x = 18;
int n = arr.Length;
int index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1)
Console.WriteLine("Element found at index " +
index);
else
Console.WriteLine("Element not found.");
}
}
// This code is contributed by equbalzeeshan
java 描述语言
<script>
// Javascript program to implement Interpolation Search
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
function interpolationSearch(arr, lo, hi, x){
let pos;
// Since array is sorted, an element present
// in array must be in range defined by corner
if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
// Probing the position with keeping
// uniform distribution in mind.
pos = lo + Math.floor(((hi - lo) / (arr[hi] - arr[lo])) * (x - arr[lo]));;
// Condition of target found
if (arr[pos] == x){
return pos;
}
// If x is larger, x is in right sub array
if (arr[pos] < x){
return interpolationSearch(arr, pos + 1, hi, x);
}
// If x is smaller, x is in left sub array
if (arr[pos] > x){
return interpolationSearch(arr, lo, pos - 1, x);
}
}
return -1;
}
// Driver Code
let arr = [10, 12, 13, 16, 18, 19, 20, 21,
22, 23, 24, 33, 35, 42, 47];
let n = arr.length;
// Element to be searched
let x = 18
let index = interpolationSearch(arr, 0, n - 1, x);
// If element was found
if (index != -1){
document.write(`Element found at index ${index}`)
}else{
document.write("Element not found");
}
// This code is contributed by _saurabh_jaiswal
</script>
Output
Element found at index 4
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