2.7 KiB
2.7 KiB
| title | localeTitle |
|---|---|
| Exponential Search | 指数搜索 |
指数搜索
指数搜索也称为手指搜索，通过每次迭代跳过2^i元素来搜索排序数组中的元素，其中i表示 循环控制变量的值，然后验证在最后一次跳转和当前跳转之间是否存在搜索元素
复杂性最坏的情况
为O（log（N）） 由于名称经常混淆，因此算法的命名不是因为时间复杂性。 该名称是由于算法跳过具有等于2的指数的步骤的元素而产生的
作品
- 一次跳转数组
2^i元素，搜索条件Array[2^(i-1)] < valueWanted < Array[2^i]。如果2^i大于数组的长度，则将上限设置为数组的长度。 - 在
Array[2^(i-1)]和Array[2^i]之间进行二进制搜索
码
// C++ program to find an element x in a
// sorted array using Exponential search.
#include <bits/stdc++.h>
using namespace std;
int binarySearch(int arr[], int, int, int);
// Returns position of first ocurrence of
// x in array
int exponentialSearch(int arr[], int n, int x)
{
// If x is present at firt location itself
if (arr[0] == x)
return 0;
// Find range for binary search by
// repeated doubling
int i = 1;
while (i < n && arr[i] <= x)
i = i*2;
// Call binary search for the found range.
return binarySearch(arr, i/2, min(i, n), x);
}
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is
// present, otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l)
{
int mid = l + (r - l)/2;
// If the element is present at the middle
// itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then it
// can only be present n left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid-1, x);
// Else the element can only be present
// in right subarray
return binarySearch(arr, mid+1, r, x);
}
// We reach here when element is not present
// in array
return -1;
}
int main(void)
{
int arr[] = {2, 3, 4, 10, 40};
int n = sizeof(arr)/ sizeof(arr[0]);
int x = 10;
int result = exponentialSearch(arr, n, x);
(result == -1)? printf("Element is not present in array")
: printf("Element is present at index %d", result);
return 0;
}