95 lines
		
	
	
		
			2.7 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
		
		
			
		
	
	
			95 lines
		
	
	
		
			2.7 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
|   | --- | |||
|  | title: Exponential Search | |||
|  | localeTitle: 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
 | |||
|  | 
 | |||
|  | 1.  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。 | |||
|  | 2.  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;  | |||
|  |  }  | |||
|  | ``` | |||
|  | 
 | |||
|  | # 更多信息
 | |||
|  | 
 | |||
|  | *   [维基百科](https://en.wikipedia.org/wiki/Exponential_search) | |||
|  |      | |||
|  | *   [GeeksForGeeks](https://www.geeksforgeeks.org/exponential-search/) | |||
|  |      | |||
|  | 
 | |||
|  | # 积分
 | |||
|  | 
 | |||
|  | [C ++实现](https://www.wikitechy.com/technology/exponential-search/) |