95 lines
2.7 KiB
Markdown
95 lines
2.7 KiB
Markdown
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---
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title: Exponential Search
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localeTitle: 指数搜索
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---
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## 指数搜索
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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
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# 复杂性最坏的情况
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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
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# 作品
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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。
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2. 在`Array[2^(i-1)]`和`Array[2^i]`之间进行二进制搜索
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# 码
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```
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// C++ program to find an element x in a
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// sorted array using Exponential search.
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#include <bits/stdc++.h>
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using namespace std;
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int binarySearch(int arr[], int, int, int);
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// Returns position of first ocurrence of
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// x in array
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int exponentialSearch(int arr[], int n, int x)
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{
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// If x is present at firt location itself
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if (arr[0] == x)
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return 0;
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// Find range for binary search by
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// repeated doubling
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int i = 1;
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while (i < n && arr[i] <= x)
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i = i*2;
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// Call binary search for the found range.
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return binarySearch(arr, i/2, min(i, n), x);
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}
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// A recursive binary search function. It returns
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// location of x in given array arr[l..r] is
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// present, otherwise -1
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int binarySearch(int arr[], int l, int r, int x)
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{
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if (r >= l)
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{
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int mid = l + (r - l)/2;
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// If the element is present at the middle
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// itself
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if (arr[mid] == x)
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return mid;
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// If element is smaller than mid, then it
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// can only be present n left subarray
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if (arr[mid] > x)
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return binarySearch(arr, l, mid-1, x);
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// Else the element can only be present
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// in right subarray
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return binarySearch(arr, mid+1, r, x);
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}
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// We reach here when element is not present
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// in array
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return -1;
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}
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int main(void)
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{
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int arr[] = {2, 3, 4, 10, 40};
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int n = sizeof(arr)/ sizeof(arr[0]);
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int x = 10;
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int result = exponentialSearch(arr, n, x);
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(result == -1)? printf("Element is not present in array")
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: printf("Element is present at index %d", result);
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return 0;
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}
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```
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# 更多信息
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* [维基百科](https://en.wikipedia.org/wiki/Exponential_search)
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* [GeeksForGeeks](https://www.geeksforgeeks.org/exponential-search/)
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# 积分
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[C ++实现](https://www.wikitechy.com/technology/exponential-search/)
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