在旋转排序数组中找到旋转计数 _数组

本文概述

建议：在继续解决方案之前, 请先在"实践"上解决它。
C ++
Java
Python3
C#
的PHP
C ++
Java
Python3
C#
的PHP

考虑以递增顺序排序的一组不同数字。阵列(顺时针)旋转了k次。给定这样一个数组, 找到k的值。
例子：

Input : arr[] = {15, 18, 2, 3, 6, 12}Output: 2Explanation : Initial array must be {2, 3, 6, 12, 15, 18}. We get the given array after rotating the initial array twice.Input : arr[] = {7, 9, 11, 12, 5}Output: 4Input: arr[] = {7, 9, 11, 12, 15}; Output: 0

推荐：请在"实践首先, 在继续解决方案之前。
方法1(使用线性搜索)
如果仔细研究示例, 我们会发现转数等于最小元素的索引。一个简单的线性解决方案是找到最小元素并返回其索引。以下是该想法的C ++实现。
C ++

// C++ program to find number of rotations // in a sorted and rotated array. #include< bits/stdc++.h> using namespace std; // Returns count of rotations for an array which // is first sorted in ascending order, then rotated int countRotations( int arr[], int n) { // We basically find index of minimum // element int min = arr[0], min_index; for ( int i=0; i< n; i++) { if (min > arr[i]) { min = arr[i]; min_index = i; } } return min_index; }// Driver code int main() { int arr[] = {15, 18, 2, 3, 6, 12}; int n = sizeof (arr)/ sizeof (arr[0]); cout < < countRotations(arr, n); return 0; }

Java

// Java program to find number of // rotations in a sorted and rotated // array. import java.util.*; import java.lang.*; import java.io.*; class LinearSearch { // Returns count of rotations for an // array which is first sorted in // ascending order, then rotated static int countRotations( int arr[], int n) { // We basically find index of minimum // element int min = arr[ 0 ], min_index = - 1 ; for ( int i = 0 ; i < n; i++) { if (min > arr[i]) { min = arr[i]; min_index = i; } } return min_index; }// Driver program to test above functions public static void main (String[] args) { int arr[] = { 15 , 18 , 2 , 3 , 6 , 12 }; int n = arr.length; System.out.println(countRotations(arr, n)); } } // This code is contributed by Chhavi

Python3

# Python3 program to find number # of rotations in a sorted and # rotated array.# Returns count of rotations for # an array which is first sorted # in ascending order, then rotated def countRotations(arr, n):# We basically find index # of minimum element min = arr[ 0 ] for i in range ( 0 , n):if ( min > arr[i]):min = arr[i] min_index = ireturn min_index; # Driver code arr = [ 15 , 18 , 2 , 3 , 6 , 12 ] n = len (arr) print (countRotations(arr, n))# This code is contributed by Smitha Dinesh Semwal

// c# program to find number of // rotations in a sorted and rotated // array. using System; class LinearSearch { // Returns count of rotations for an // array which is first sorted in // ascending order, then rotated static int countRotations( int []arr, int n) { // We basically find index of minimum // element int min = arr[0], min_index = -1; for ( int i = 0; i < n; i++) { if (min > arr[i]) { min = arr[i]; min_index = i; } } return min_index; }// Driver program to test above functions public static void Main () { int []arr = {15, 18, 2, 3, 6, 12}; int n = arr.Length; Console.WriteLine(countRotations(arr, n)); } } // This code is contributed by vt_m.

的PHP

< ?php // PHP program to find number // of rotations in a sorted // and rotated array.// Returns count of rotations // for an array which is first // sorted in ascending order, // then rotated function countRotations( $arr , $n ) { // We basically find index // of minimum element $min = $arr [0]; $min_index ; for ( $i = 0; $i < $n ; $i ++) { if ( $min > $arr [ $i ]) { $min = $arr [ $i ]; $min_index = $i ; } } return $min_index ; }// Driver code $arr = array (15, 18, 2, 3, 6, 12); $n = sizeof( $arr ); echo countRotations( $arr , $n ); // This code is contributed // by ajit ?>

输出如下：

2

时间复杂度：
上)
辅助空间：
O(1)
方法2(高效使用
二元搜寻
)
在这里我们也找到最小元素的索引, 但是使用二进制搜索。这个想法基于以下事实：

最小元素是前一个大于它的唯一元素。如果没有先前的元素, 则没有旋转(第一个元素为最小)。我们通过将此条件与第(mid-1)个和第(mid + 1)个元素进行比较来检查此条件。
如果最小元素不在中间(既不是中也不是中+1), 则最小元素位于左半部分或右半部分。
1. 如果中间元素小于最后一个元素, 则最小元素位于左半部分
2. 其他最小元素位于右半部分。

【在旋转排序数组中找到旋转计数】下面是从中获取的实现
这里
.
C ++

// Binary Search based C++ program to find number // of rotations in a sorted and rotated array. #include< bits/stdc++.h> using namespace std; // Returns count of rotations for an array which // is first sorted in ascending order, then rotated int countRotations( int arr[], int low, int high) { // This condition is needed to handle the case // when the array is not rotated at all if (high < low) return 0; // If there is only one element left if (high == low) return low; // Find mid int mid = low + (high - low)/2; /*(low + high)/2; */// Check if element (mid+1) is minimum element. // Consider the cases like {3, 4, 5, 1, 2} if (mid < high & & arr[mid+1] < arr[mid]) return (mid+1); // Check if mid itself is minimum element if (mid > low & & arr[mid] < arr[mid - 1]) return mid; // Decide whether we need to go to left half or // right half if (arr[high] > arr[mid]) return countRotations(arr, low, mid-1); return countRotations(arr, mid+1, high); }// Driver code int main() { int arr[] = {15, 18, 2, 3, 6, 12}; int n = sizeof (arr)/ sizeof (arr[0]); cout < < countRotations(arr, 0, n-1); return 0; }

Java

// Java program to find number of // rotations in a sorted and rotated // array. import java.util.*; import java.lang.*; import java.io.*; class BinarySearch { // Returns count of rotations for an array // which is first sorted in ascending order, // then rotated static int countRotations( int arr[], int low, int high) { // This condition is needed to handle // the case when array is not rotated // at all if (high < low) return 0 ; // If there is only one element left if (high == low) return low; // Find mid // /*(low + high)/2; */ int mid = low + (high - low)/ 2 ; // Check if element (mid+1) is minimum // element. Consider the cases like // {3, 4, 5, 1, 2} if (mid < high & & arr[mid+ 1 ] < arr[mid]) return (mid + 1 ); // Check if mid itself is minimum element if (mid > low & & arr[mid] < arr[mid - 1 ]) return mid; // Decide whether we need to go to left // half or right half if (arr[high] > arr[mid]) return countRotations(arr, low, mid - 1 ); return countRotations(arr, mid + 1 , high); }// Driver program to test above functions public static void main (String[] args) { int arr[] = { 15 , 18 , 2 , 3 , 6 , 12 }; int n = arr.length; System.out.println(countRotations(arr, 0 , n- 1 )); } } // This code is contributed by Chhavi

Python3

# Binary Search based Python3 # program to find number of # rotations in a sorted and # rotated array.# Returns count of rotations for # an array which is first sorted # in ascending order, then rotated def countRotations(arr, low, high):# This condition is needed to # handle the case when array # is not rotated at all if (high < low): return 0# If there is only one # element left if (high = = low): return low# Find mid # (low + high)/2 mid = low + (high - low) / 2 ; mid = int (mid)# Check if element (mid+1) is # minimum element. Consider # the cases like {3, 4, 5, 1, 2} if (mid < high and arr[mid + 1 ] < arr[mid]): return (mid + 1 )# Check if mid itself is # minimum element if (mid > low and arr[mid] < arr[mid - 1 ]): return mid# Decide whether we need to go # to left half or right half if (arr[high] > arr[mid]): return countRotations(arr, low, mid - 1 ); return countRotations(arr, mid + 1 , high)# Driver code arr = [ 15 , 18 , 2 , 3 , 6 , 12 ] n = len (arr) print (countRotations(arr, 0 , n - 1 ))# This code is contributed by Smitha Dinesh Semwal

// C# program to find number of // rotations in a sorted and rotated // array. using System; class BinarySearch { // Returns count of rotations for an array // which is first sorted in ascending order, // then rotated static int countRotations( int []arr, int low, int high) { // This condition is needed to handle // the case when array is not rotated // at all if (high < low) return 0; // If there is only one element left if (high == low) return low; // Find mid // /*(low + high)/2; */ int mid = low + (high - low)/2; // Check if element (mid+1) is minimum // element. Consider the cases like // {3, 4, 5, 1, 2} if (mid < high & & arr[mid+1] < arr[mid]) return (mid + 1); // Check if mid itself is minimum element if (mid > low & & arr[mid] < arr[mid - 1]) return mid; // Decide whether we need to go to left // half or right half if (arr[high] > arr[mid]) return countRotations(arr, low, mid - 1); return countRotations(arr, mid + 1, high); }// Driver program to test above functions public static void Main () { int []arr = {15, 18, 2, 3, 6, 12}; int n = arr.Length; Console.WriteLine(countRotations(arr, 0, n-1)); } } // This code is contributed by vt_m.

的PHP

< ?php // Binary Search based PHP // program to find number // of rotations in a sorted // and rotated array.// Returns count of rotations // for an array which is // first sorted in ascending // order, then rotated function countRotations( $arr , $low , $high ) { // This condition is needed // to handle the case when // array is not rotated at all if ( $high < $low ) return 0; // If there is only // one element left if ( $high == $low ) return $low ; // Find mid $mid = $low + ( $high - $low ) / 2; // Check if element (mid+1) // is minimum element. Consider // the cases like {3, 4, 5, 1, 2} if ( $mid < $high & & $arr [ $mid + 1] < $arr [ $mid ]) return (int)( $mid + 1); // Check if mid itself // is minimum element if ( $mid > $low & & $arr [ $mid ] < $arr [ $mid - 1]) return (int)( $mid ); // Decide whether we need // to go to left half or // right half if ( $arr [ $high ] > $arr [ $mid ]) return countRotations( $arr , $low , $mid - 1); return countRotations( $arr , $mid + 1, $high ); }// Driver code $arr = array (15, 18, 2, 3, 6, 12); $n = sizeof( $arr ); echo countRotations( $arr , 0, $n - 1); // This code is contributed bu ajit ?>

输出如下：