Surūšiuotas n vienodai paskirstytų reikšmių masyvas arr[] parašykite funkciją tam, kad masyve ieškotumėte konkretaus elemento x.
Linijinė paieška suranda elementą per O(n) laiką Peršokti paieška užtrunka O(n) laiko ir Dvejetainė paieška užtrunka O(log n) laiko.
Interpoliacijos paieška yra patobulinta Dvejetainė paieška atvejams, kai surūšiuoto masyvo reikšmės yra tolygiai paskirstytos. Interpoliacija sukuria naujus duomenų taškus atskiros žinomų duomenų taškų rinkinio diapazone. Dvejetainė paieška visada eina į vidurinį elementą patikrinti. Kita vertus, interpoliacijos paieška gali vykti į skirtingas vietas, atsižvelgiant į ieškomo rakto reikšmę. Pavyzdžiui, jei rakto reikšmė yra arčiau paskutinio elemento, interpoliacijos paieška greičiausiai pradės paiešką link galo.
Norėdami rasti ieškomą poziciją, ji naudoja šią formulę.
// Formulės idėja yra grąžinti didesnę poz
// kai elementas, kurio reikia ieškoti, yra arčiau arr[hi]. Ir
// mažesnė reikšmė, kai arčiau arr[lo]
arr[] ==> Masyvas, kuriame reikia ieškoti elementų
x ==> Elementas, kurio reikia ieškoti
kas yra Fredis Merkurislo ==> Pradinis indeksas arr[]
labas ==> Indeksas baigiasi arr[]
po = +
nfa pavyzdžiai
Yra daug skirtingų interpoliacijos metodų, vienas iš jų yra žinomas kaip tiesinė interpoliacija. Tiesinė interpoliacija paima du duomenų taškus, kuriuos laikome (x1y1) ir (x2y2), o formulė yra : taške (xy).
Šis algoritmas veikia taip, kaip ieškome žodžio žodyne. Interpoliacijos paieškos algoritmas pagerina dvejetainės paieškos algoritmą. Reikšmės radimo formulė yra tokia: K = >K yra konstanta, kuri naudojama paieškos erdvei susiaurinti. Dvejetainės paieškos atveju šios konstantos reikšmė yra: K=(žemas+aukštas)/2.
eilutė c++
Pos formulę galima gauti taip.
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 lohi 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])
Algoritmas
Likusi interpoliavimo algoritmo dalis yra tokia pati, išskyrus aukščiau pateiktą skaidymo logiką.
- 1 veiksmas: Cikle apskaičiuokite „pos“ reikšmę naudodami zondo padėties formulę.
- 2 veiksmas: Jei tai atitinka, grąžinkite elemento indeksą ir išeikite.
- 3 veiksmas: Jei elementas yra mažesnis nei arr[pos], apskaičiuokite kairiojo pomasyvo zondo padėtį. Kitu atveju apskaičiuokite tą patį dešiniajame pomasyve.
- 4 veiksmas: Kartokite tol, kol bus rastas atitikmuo arba antrinis masyvas sumažės iki nulio.
Žemiau pateikiamas algoritmo įgyvendinimas.
// C++ program to implement interpolation // search with recursion #include
// C program to implement interpolation search // with recursion #include
// 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
# 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# 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
<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>
// PHP program to implement $erpolation search // with recursion // If x is present in arr[0..n-1] then returns // index of it else returns -1. function interpolationSearch($arr $lo $hi $x) { // Since array is sorted an element present // in array must be in range defined by corner if ($lo <= $hi && $x >= $arr[$lo] && $x <= $arr[$hi]) { // Probing the position with keeping // uniform distribution in mind. $pos = (int)($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 // Array of items on which search will // be conducted. $arr = array(10 12 13 16 18 19 20 21 22 23 24 33 35 42 47); $n = sizeof($arr); $x = 47; // Element to be searched $index = interpolationSearch($arr 0 $n - 1 $x); // If element was found if ($index != -1) echo 'Element found at index '.$index; else echo 'Element not found.'; return 0; #This code is contributed by Susobhan Akhuli ?> Išvestis
Element found at index 4
Laiko sudėtingumas: O(log2(log2n)) vidutiniam atvejui ir O(n) blogiausiam atvejui
Pagalbinės erdvės sudėtingumas: O(1)
Kitas požiūris:
Tai yra iteracijos metodas interpoliacijos paieškai.
- 1 veiksmas: Cikle apskaičiuokite „pos“ reikšmę naudodami zondo padėties formulę.
- 2 veiksmas: Jei tai atitinka, grąžinkite elemento indeksą ir išeikite.
- 3 veiksmas: Jei elementas yra mažesnis nei arr[pos], apskaičiuokite kairiojo pomasyvo zondo padėtį. Kitu atveju apskaičiuokite tą patį dešiniajame pomasyve.
- 4 veiksmas: Kartokite tol, kol bus rastas atitikmuo arba antrinis masyvas sumažės iki nulio.
Žemiau pateikiamas algoritmo įgyvendinimas.
runas in powershellC++
// C++ program to implement interpolation search by using iteration approach #include
// 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; 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.'); } }
# Python equivalent of above C++ code # Python program to implement interpolation search by using iteration approach def interpolationSearch(arr n x): # Find indexes of two corners low = 0 high = (n - 1) # Since array is sorted an element present # in array must be in range defined by corner while low <= high and x >= arr[low] and x <= arr[high]: if low == high: if arr[low] == x: return low; return -1; # Probing the position with keeping # uniform distribution in mind. pos = int(low + (((float(high - low)/( arr[high] - arr[low])) * (x - arr[low])))) # Condition of target found if arr[pos] == x: return pos # If x is larger x is in upper part if arr[pos] < x: low = pos + 1; # If x is smaller x is in lower part else: high = pos - 1; return -1 # Main function if __name__ == '__main__': # Array of items on whighch search will # be conducted. arr = [10 12 13 16 18 19 20 21 22 23 24 33 35 42 47] n = len(arr) x = 18 # Element to be searched index = interpolationSearch(arr n x) # If element was found if index != -1: print ('Element found at index'index) else: print ('Element not found')
// C# program to implement interpolation search by using // iteration approach using System; class Program { // Interpolation Search function static int InterpolationSearch(int[] arr int n int x) { int low = 0; int high = n - 1; while (low <= high && x >= arr[low] && x <= arr[high]) { if (low == high) { if (arr[low] == x) return low; return -1; } int pos = low + (int)(((float)(high - low) / (arr[high] - arr[low])) * (x - arr[low])); if (arr[pos] == x) return pos; if (arr[pos] < x) low = pos + 1; else high = pos - 1; } return -1; } // Main function static void Main(string[] args) { int[] arr = {10 12 13 16 18 19 20 21 22 23 24 33 35 42 47}; int n = arr.Length; int x = 18; int index = InterpolationSearch(arr n x); if (index != -1) Console.WriteLine('Element found at index ' + index); else Console.WriteLine('Element not found'); } } // This code is contributed by Susobhan Akhuli
// JavaScript program to implement interpolation search by using iteration approach function interpolationSearch(arr n x) { // Find indexes of two corners let low = 0; let high = n - 1; // Since array is sorted an element present // in array must be in range defined by corner while (low <= high && x >= arr[low] && x <= arr[high]) { if (low == high) { if (arr[low] == x) { return low; } return -1; } // Probing the position with keeping // uniform distribution in mind. let pos = Math.floor(low + (((high - low) / (arr[high] - arr[low])) * (x - arr[low]))); // Condition of target found if (arr[pos] == x) { return pos; } // If x is larger x is in upper part if (arr[pos] < x) { low = pos + 1; } // If x is smaller x is in lower part else { high = pos - 1; } } return -1; } // Main function let arr = [10 12 13 16 18 19 20 21 22 23 24 33 35 42 47]; let n = arr.length; let x = 18; // Element to be searched let index = interpolationSearch(arr n x); // If element was found if (index != -1) { console.log('Element found at index' index); } else { console.log('Element not found'); }
Išvestis
Element found at index 4
Laiko sudėtingumas: O(log2(log2 n)) vidutiniam atvejui ir O(n) blogiausiam atvejui
Pagalbinės erdvės sudėtingumas: O(1)