# 以最大最小形式重新排列數組| 系列 1 > 原文： [https://www.geeksforgeeks.org/rearrange-array-maximum-minimum-form/](https://www.geeksforgeeks.org/rearrange-array-maximum-minimum-form/) 給定一個帶正整數的排序數組，請交替排列數組，即第一個元素應為最大值，第二個最小值，第三個第二個最大值，第四個第二個最小值等。 **示例**： > **輸入**：`arr[] = {1, 2, 3, 4, 5, 6, 7}` > **輸出**：`arr[] = {7, 1, 6, 2 , 5, 3, 4}` > > **輸入**：`arr[] = {1, 2, 3, 4, 5, 6}` > **輸出**：`arr[] = {6, 1, 5, 2, 4, 3}` 預期時間復雜度：`O(n)`。 這個想法是使用一個輔助數組。 我們維護兩個指針，一個指向最左或最小元素，另一個指向最右或最大元素。 我們將兩個指針都指向彼此更多，或者將這些指針處的元素復制到輔助數組。 最后，我們將輔助數組復制回原始數組。 下圖是上述方法的模擬：  下面是上述方法的實現： ## C++ ```cpp // C++ program to rearrange an array in minimum // maximum form #include <bits/stdc++.h> using namespace std; // Prints max at first position, min at second position // second max at third position, second min at fourth // position and so on. void rearrange(int arr[], int n) { ????// Auxiliary array to hold modified array ????int temp[n]; ????// Indexes of smallest and largest elements ????// from remaining array. ????int small=0, large=n-1; ????// To indicate whether we need to copy rmaining ????// largest or remaining smallest at next position ????int flag = true; ????// Store result in temp[] ????for (int i=0; i<n; i++) ????{ ????????if (flag) ????????????temp[i] = arr[large--]; ????????else ????????????temp[i] = arr[small++]; ????????flag = !flag; ????} ????// Copy temp[] to arr[] ????for (int i=0; i<n; i++) ????????arr[i] = temp[i]; } // Driver program to test above function int main() { ????int arr[] = {1, 2, 3, 4, 5, 6}; ????int n = sizeof(arr)/sizeof(arr[0]); ????cout << "Original Arrayn"; ????for (int i=0; i<n; i++) ????????cout << arr[i] << " "; ????rearrange(arr, n); ????cout << "nModified Arrayn"; ????for (int i=0; i<n; i++) ????????cout << arr[i] << " "; ????return 0; } ``` ## Java ```java // Java program to rearrange an array in minimum // maximum form import java.util.Arrays; public class GFG { ????// Prints max at first position, min at second position ????// second max at third position, second min at fourth ????// position and so on. ????static void rearrange(int[] arr, int n) ????{ ????????// Auxiliary array to hold modified array ????????int temp[] = new int[n]; ????????// Indexes of smallest and largest elements ????????// from remaining array. ????????int small=0, large=n-1; ????????// To indicate whether we need to copy rmaining ????????// largest or remaining smallest at next position ????????boolean flag = true; ????????// Store result in temp[] ????????for (int i=0; i<n; i++) ????????{ ????????????if (flag) ????????????????temp[i] = arr[large--]; ????????????else ????????????????temp[i] = arr[small++]; ????????????flag = !flag; ????????} ????????// Copy temp[] to arr[] ????????arr = temp.clone(); ????} ????// Driver method to test the above function ????public static void main(String[] args)? ????{ ????????int arr[] = new int[]{1, 2, 3, 4, 5, 6}; ????????System.out.println("Original Array "); ????????System.out.println(Arrays.toString(arr)); ????????rearrange(arr,arr.length); ????????System.out.println("Modified Array "); ????????System.out.println(Arrays.toString(arr)); ????} } ``` ## Python ``` # Python program to rearrange an array in minimum # maximum form # Prints max at first position, min at second position # second max at third position, second min at fourth # position and so on. def rearrange(arr, n): ????# Auxiliary array to hold modified array ????temp = n*[None] ????# Indexes of smallest and largest elements ????# from remaining array. ????small,large =0,n-1 ????# To indicate whether we need to copy rmaining ????# largest or remaining smallest at next position ????flag = True ????# Store result in temp[] ????for i in range(n): ????????if flag is True: ????????????temp[i] = arr[large] ????????????large -= 1 ????????else: ????????????temp[i] = arr[small] ????????????small += 1 ????????flag = bool(1-flag) ????# Copy temp[] to arr[] ????for i in range(n): ????????arr[i] = temp[i] ????return arr # Driver program to test above function arr = [1, 2, 3, 4, 5, 6] n = len(arr) print("Original Array") print(arr) print("Modified Array") print(rearrange(arr, n)) # This code is contributed by Pratik Chhajer ``` ## C# ```cs // C# program to rearrange? // an array in minimum? // maximum form using System; class GFG { // Prints max at first position,? // min at second position second? // max at third position, second? // min at fourth position and so on. static void rearrage(int[] arr,? ?????????????????????int n) { ????// Auxiliary array to? ????// hold modified array ????int []temp = new int[n]; ????// Indexes of smallest? ????// and largest elements ????// from remaining array. ????int small = 0, large = n - 1; ????// To indicate whether we? ????// need to copy rmaining ????// largest or remaining? ????// smallest at next position ????bool flag = true; ????// Store result in temp[] ????for (int i = 0; i < n; i++) ????{ ????????if (flag) ????????????temp[i] = arr[large--]; ????????else ????????????temp[i] = arr[small++]; ????????flag = !flag; ????} ????// Copy temp[] to arr[] ????for (int i = 0; i < n; i++) ????????arr[i] = temp[i]; } // Driver Code static void Main() { ????int[] arr = {1, 2, 3, 4,? ?????????????????5, 6}; ????Console.WriteLine("Original Array"); ????for(int i = 0; i < arr.Length; i++) ????????Console.Write(arr[i] + " "); ????rearrage(arr, arr.Length);????? ????Console.WriteLine("\nModified Array"); ????for(int i = 0; i < arr.Length; i++) ????????Console.Write(arr[i] + " "); } } // This code is contributed // by Sam007 ``` ## PHP ```php <?php // PHP program to rearrange an array in? // minimum-maximum form // Prints max at first position, min at? // second position second max at third? // position, second min at fourth // position and so on. function rearrange(&$arr, $n) { ????// Auxiliary array to hold modified array ????$temp = array(); ????// Indexes of smallest and largest elements ????// from remaining array. ????$small = 0; ????$large = $n - 1; ????// To indicate whether we need to copy? ????// remaining largest or remaining smallest ????// at next position ????$flag = true; ????// Store result in temp[] ????for ($i = 0; $i < $n; $i++) ????{ ????????if ($flag) ????????????$temp[$i] = $arr[$large--]; ????????else ????????????$temp[$i] = $arr[$small++]; ????????$flag = !$flag; ????} ????// Copy temp[] to arr[] ????for ($i = 0; $i < $n; $i++) ????????$arr[$i] = $temp[$i]; } // Driver Code $arr = array(1, 2, 3, 4, 5, 6); $n = count($arr); echo "Original Arrayn\n"; for ($i = 0; $i < $n; $i++) ????echo $arr[$i] . " "; rearrange($arr, $n); echo "\nModified Arrayn\n"; for ($i = 0; $i < $n; $i++) ????echo $arr[$i] . " "; // This code is contributed by // Rajput-Ji ?> ``` **輸出**： ``` Original Array 1 2 3 4 5 6 Modified Array 6 1 5 2 4 3 ``` **時間復雜度**：`O(n)` **輔助空間**：`O(n)` **練習**：如果不允許多余的空間，如何解決此問題？ [**以最大最小形式重新排列數組| 系列 2（`O(1)`多余的空間）**](https://www.geeksforgeeks.org/rearrange-array-maximum-minimum-form-set-2-o1-extra-space/)