假设您有一个无穷大的排序数组，您将如何搜索该数组中的元素?
资料来源：Amazon Interview Experience。
由于数组已排序，因此首先想到的是二进制搜索，但是这里的问题是我们不知道数组的大小。
如果数组是无限的，则意味着我们没有适当的界限来应用二进制搜索。因此，为了找到键的位置，我们首先找到边界，然后应用二进制搜索算法。
让low指向数组的第一个元素，high指向数组的第二个元素，现在将键与高索引元素进行比较，
->如果它大于高索引元素，则将高索引复制到低索引中，并将高索引加倍。
->如果较小，则对找到的高索引和低索引进行二进制搜索。
以下是上述算法的实现
C++
// C++ program to demonstrate working of an algorithm that finds
// an element in an array of infinite size
#include
using namespace std;
// Simple binary search algorithm
int binarySearch(int arr[], int l, int r, int x)
{
if (r>=l)
{
int mid = l + (r - l)/2;
if (arr[mid] == x)
return mid;
if (arr[mid] > x)
return binarySearch(arr, l, mid-1, x);
return binarySearch(arr, mid+1, r, x);
}
return -1;
}
// function takes an infinite size array and a key to be
// searched and returns its position if found else -1.
// We don't know size of arr[] and we can assume size to be
// infinite in this function.
// NOTE THAT THIS FUNCTION ASSUMES arr[] TO BE OF INFINITE SIZE
// THEREFORE, THERE IS NO INDEX OUT OF BOUND CHECKING
int findPos(int arr[], int key)
{
int l = 0, h = 1;
int val = arr[0];
// Find h to do binary search
while (val < key)
{
l = h; // store previous high
h = 2*h; // double high index
val = arr[h]; // update new val
}
// at this point we have updated low and
// high indices, Thus use binary search
// between them
return binarySearch(arr, l, h, key);
}
// Driver program
int main()
{
int arr[] = {3, 5, 7, 9, 10, 90, 100, 130,
140, 160, 170};
int ans = findPos(arr, 10);
if (ans==-1)
cout << "Element not found";
else
cout << "Element found at index " << ans;
return 0;
} Java
// Java program to demonstrate working of
// an algorithm that finds an element in an
// array of infinite size
class Test
{
// Simple binary search algorithm
static int binarySearch(int arr[], int l, int r, int x)
{
if (r>=l)
{
int mid = l + (r - l)/2;
if (arr[mid] == x)
return mid;
if (arr[mid] > x)
return binarySearch(arr, l, mid-1, x);
return binarySearch(arr, mid+1, r, x);
}
return -1;
}
// Method takes an infinite size array and a key to be
// searched and returns its position if found else -1.
// We don't know size of arr[] and we can assume size to be
// infinite in this function.
// NOTE THAT THIS FUNCTION ASSUMES arr[] TO BE OF INFINITE SIZE
// THEREFORE, THERE IS NO INDEX OUT OF BOUND CHECKING
static int findPos(int arr[],int key)
{
int l = 0, h = 1;
int val = arr[0];
// Find h to do binary search
while (val < key)
{
l = h; // store previous high
//check that 2*h doesn't exceeds array
//length to prevent ArrayOutOfBoundException
if(2*h < arr.length-1)
h = 2*h;
else
h = arr.length-1;
val = arr[h]; // update new val
}
// at this point we have updated low
// and high indices, thus use binary
// search between them
return binarySearch(arr, l, h, key);
}
// Driver method to test the above function
public static void main(String[] args)
{
int arr[] = new int[]{3, 5, 7, 9, 10, 90,
100, 130, 140, 160, 170};
int ans = findPos(arr,10);
if (ans==-1)
System.out.println("Element not found");
else
System.out.println("Element found at index " + ans);
}
}Python
# Python Program to demonstrate working of an algorithm that finds
# an element in an array of infinite size
# Binary search algorithm implementation
def binary_search(arr,l,r,x):
if r >= l:
mid = l+(r-l)/2
if arr[mid] == x:
return mid
if arr[mid] > x:
return binary_search(arr,l,mid-1,x)
return binary_search(arr,mid+1,r,x)
return -1
# function takes an infinite size array and a key to be
# searched and returns its position if found else -1.
# We don't know size of a[] and we can assume size to be
# infinite in this function.
# NOTE THAT THIS FUNCTION ASSUMES a[] TO BE OF INFINITE SIZE
# THEREFORE, THERE IS NO INDEX OUT OF BOUND CHECKING
def findPos(a, key):
l, h, val = 0, 1, arr[0]
# Find h to do binary search
while val < key:
l = h #store previous high
h = 2*h #double high index
val = arr[h] #update new val
# at this point we have updated low and high indices,
# thus use binary search between them
return binary_search(a, l, h, key)
# Driver function
arr = [3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170]
ans = findPos(arr,10)
if ans == -1:
print "Element not found"
else:
print"Element found at index",ans
# This code is contributed by __Devesh Agrawal__C#
// C# program to demonstrate working of an
// algorithm that finds an element in an
// array of infinite size
using System;
class GFG {
// Simple binary search algorithm
static int binarySearch(int []arr, int l,
int r, int x)
{
if (r >= l)
{
int mid = l + (r - l)/2;
if (arr[mid] == x)
return mid;
if (arr[mid] > x)
return binarySearch(arr, l,
mid-1, x);
return binarySearch(arr, mid+1,
r, x);
}
return -1;
}
// Method takes an infinite size array
// and a key to be searched and returns
// its position if found else -1. We
// don't know size of arr[] and we can
// assume size to be infinite in this
// function.
// NOTE THAT THIS FUNCTION ASSUMES
// arr[] TO BE OF INFINITE SIZE
// THEREFORE, THERE IS NO INDEX OUT
// OF BOUND CHECKING
static int findPos(int []arr,int key)
{
int l = 0, h = 1;
int val = arr[0];
// Find h to do binary search
while (val < key)
{
l = h; // store previous high
h = 2 * h;// double high index
val = arr[h]; // update new val
}
// at this point we have updated low
// and high indices, thus use binary
// search between them
return binarySearch(arr, l, h, key);
}
// Driver method to test the above
// function
public static void Main()
{
int []arr = new int[]{3, 5, 7, 9,
10, 90, 100, 130, 140, 160, 170};
int ans = findPos(arr, 10);
if (ans == -1)
Console.Write("Element not found");
else
Console.Write("Element found at "
+ "index " + ans);
}
}
// This code is contributed by nitin mittal.PHP
= $l)
{
$mid = $l + ($r - $l)/2;
if ($arr[$mid] == $x)
return $mid;
if ($arr[$mid] > $x)
return binarySearch($arr, $l,
$mid - 1, $x);
return binarySearch($arr, $mid + 1,
$r, $x);
}
return -1;
}
// function takes an infinite
// size array and a key to be
// searched and returns its
// position if found else -1.
// We don't know size of arr[]
// and we can assume size to be
// infinite in this function.
// NOTE THAT THIS FUNCTION ASSUMES
// arr[] TO BE OF INFINITE SIZE
// THEREFORE, THERE IS NO INDEX
// OUT OF BOUND CHECKING
function findPos( $arr, $key)
{
$l = 0; $h = 1;
$val = $arr[0];
// Find h to do binary search
while ($val < $key)
{
// store previous high
$l = $h;
// double high index
$h = 2 * $h;
// update new val
$val = $arr[$h];
}
// at this point we have
// updated low and high
// indices, Thus use binary
// search between them
return binarySearch($arr, $l,
$h, $key);
}
// Driver Code
$arr = array(3, 5, 7, 9, 10, 90, 100,
130, 140, 160, 170);
$ans = findPos($arr, 10);
if ($ans==-1)
echo "Element not found";
else
echo "Element found at index " , $ans;
// This code is contributed by anuj_67.
?>输出：
Element found at index 4令p为要搜索元素的位置。查找高索引“ h”的步骤数为O(Log p)。 “ h”的值必须小于2 * p。 h / 2和h之间的元素数必须为O(p)。因此，二分搜索步骤的时间复杂度也为O(Log p)，总时间复杂度为2 * O(Log p)，即O(Log p)。