正数元素位于偶数位置，负数元素位于奇数位置（不保持相对顺序） · GeeksForGeeks 中文系列教程

# 正数元素位于偶数位置，负数元素位于奇数位置（不保持相对顺序） > 原文： [https://www.geeksforgeeks.org/positive-elements-even-negative-odd-positions/](https://www.geeksforgeeks.org/positive-elements-even-negative-odd-positions/) 给您一个数组，您必须编写一个程序来转换该数组，以使正元素出现在数组的偶数位置，而负元素出现在数组的奇数位置。我们必须做到这一点。正值和负值的数目可能不相等，多余的值必须保持原样。 **示例**： ``` Input : arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10} Output : arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10} Input : arr[] = {-1, 3, -5, 6, 3, 6, -7, -4, -9, 10} Output : arr[] = {3, -1, 6, -5, 3, -7, 6, -4, 10, -9} ``` 这个想法是使用[快速排序](https://www.geeksforgeeks.org/quick-sort/)的 [Hoare 分区](https://www.geeksforgeeks.org/hoares-vs-lomuto-partition-scheme-quicksort/)处理。我们采用正面和负面两个指标。我们将正指针设置在数组的开头，将负指针设置在数组的第一个位置。我们将正指针向前移动两步，直到找到负元素。同样，我们将负指针向前移动两个位置，直到在其位置找到正值。如果正负指针位于数组中，则我们将交换这些索引处的值，否则将停止执行该过程。 ## C++ ```cpp // C++ program to rearrange positive and negative // numbers #include <bits/stdc++.h> using namespace std; void rearrange(int a[], int size) { int positive = 0, negative = 1; while (true) { /* Move forward the positive pointer till negative number number not encountered */ while (positive < size && a[positive] >= 0) positive += 2; /* Move forward the negative pointer till positive number number not encountered */ while (negative < size && a[negative] <= 0) negative += 2; // Swap array elements to fix their position. if (positive < size && negative < size) swap(a[positive], a[negative]); /* Break from the while loop when any index exceeds the size of the array */ else break; } } // Driver code int main() { int arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 }; int n = (sizeof(arr) / sizeof(arr[0])); rearrange(arr, n); for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0; } ``` ## Java ```java // Java program to rearrange positive // and negative numbers import java.io.*; class GFG { static void rearrange(int a[], int size) { int positive = 0, negative = 1, temp; while (true) { /* Move forward the positive pointer till negative number number not encountered */ while (positive < size && a[positive] >= 0) positive += 2; /* Move forward the negative pointer till positive number number not encountered */ while (negative < size && a[negative] <= 0) negative += 2; // Swap array elements to fix their position. if (positive < size && negative < size) { temp = a[positive]; a[positive] = a[negative]; a[negative] = temp; } /* Break from the while loop when any index exceeds the size of the array */ else break; } } // Driver code public static void main(String args[]) { int arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}; int n = arr.length; rearrange(arr, n); for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); } } /*This code is contributed by Nikita Tiwari.*/ ``` ## Python3 ```py # Python 3 program to rearrange # positive and negative numbers def rearrange(a, size) : positive = 0 negative = 1 while (True) : # Move forward the positive # pointer till negative number # number not encountered while (positive < size and a[positive] >= 0) : positive = positive + 2 # Move forward the negative # pointer till positive number # number not encountered while (negative < size and a[negative] <= 0) : negative = negative + 2 # Swap array elements to fix # their position. if (positive < size and negative < size) : temp = a[positive] a[positive] = a[negative] a[negative] = temp # Break from the while loop when # any index exceeds the size of # the array else : break # Driver code arr =[ 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 ] n = len(arr) rearrange(arr, n) for i in range(0, n) : print(arr[i], end = " ") # This code is contributed by Nikita Tiwari. ``` ## C# ```cs // C# program to rearrange positive // and negative numbers using System; class GFG { // Function to rearrange static void rearrange(int []a, int size) { int positive = 0, negative = 1, temp; while (true) { // Move forward the positive pointer till // negative number number not encountered while (positive < size && a[positive] >= 0) positive += 2; // Move forward the negative pointer till // positive number number not encountered while (negative < size && a[negative] <= 0) negative += 2; // Swap array elements to fix their position. if (positive < size && negative < size) { temp = a[positive]; a[positive] = a[negative]; a[negative] = temp; } // Break from the while loop when any // index exceeds the size of the array else break; } } // Driver Code public static void Main(String []args) { int []arr = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}; int n = arr.Length; rearrange(arr, n); for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); } } // This code is contributed by Nitin Mittal. ``` ## PHP ```php <?php // PHP program to rearrange positive // and negative numbers function rearrange(&$a, $size) { $positive = 0; $negative = 1; while (true) { /* Move forward the positive pointer till negative number number not encountered */ while ($positive < $size && $a[$positive] >= 0) $positive += 2; /* Move forward the negative pointer till positive number number not encountered */ while ($negative < $size && $a[$negative] <= 0) $negative += 2; // Swap array elements to fix // their position. if ($positive < $size && $negative < $size) { $temp = $a[$positive]; $a[$positive] = $a[$negative]; $a[$negative] = $temp; } /* Break from the while loop when any index exceeds the size of the array */ else break; } } // Driver code $arr = array( 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 ); $n = sizeof($arr); rearrange($arr, $n); for ($i = 0; $i < $n; $i++) echo $arr[$i] ." "; // This code is contributed by ChitraNayal ?> ``` **输出**： ``` 1 -3 5 -3 6 6 7 -4 9 10 ``` 让我们在第一个示例上解释代码的工作。 `arr[] = {1, -3, 5, 6, -3, 6, 7, -4, 9, 10}` 我们声明两个变量为正负，正数指向第零个位置，负数指向第一个位置正`= 0`，负数`= 1` 在第一次迭代中，由于`a[4]`小于零且正数`= 4`，正数会将 4 个位置移至第五个位置。负数将移动 2 个位置，并以`a[3] > 0`指向第四位置。我们将交换正和负位置值，因为它们小于数组的大小。第一次迭代后，数组变为`arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}` 现在在第四位置的正点和在第三位置的负点在第二次迭代中，正值将移动 6 位，其值将比数组的大小大。负指针将向前移动两步，并将指向第五个位置。当正指针值变得大于数组大小时，我们将不执行任何交换操作并退出`while`循环。最终输出为 `arr[] = {1, -3, 5, -3, 6, 6, 7, -4, 9, 10}` **不维持相对顺序的示例**： `{-1, -2, -3, -4, -5, 6, 7, 8};` **另一种方法：-** 的想法是找到一个不正确位置的正/负元素（即，在奇数位置为正，在偶数位置为负），然后找到符号相反的元素，即在剩余数组中的错误位置，然后交换这两个元素。这是上述想法的实现。 ## C++ ``` // C++ program to rearrange positive // and negative numbers #include<iostream> using namespace std; // Swap function void swap(int* a, int i , int j) { int temp = a[i]; a[i] = a[j]; a[j] = temp; return ; } // Print array function void printArray(int* a, int n) { for(int i = 0; i < n; i++) cout << a[i] << " "; cout << endl; return ; } // Driver code int main() { int arr[] = { 1, -3, 5, 6, -3, 6, 7, -4, 9, 10 }; int n = sizeof(arr)/sizeof(arr[0]); //before modification printArray(arr, n); for(int i = 0; i < n; i++) { if(arr[i] >= 0 && i % 2 == 1) { // out of order positive element for(int j = i + 1; j < n; j++) { if(arr[j] < 0 && j % 2 == 0) { // find out of order negative // element in remaining array swap(arr, i, j); break ; } } } else if(arr[i] < 0 && i % 2 == 0) { // out of order negative element for(int j = i + 1; j < n; j++) { if(arr[j] >= 0 && j % 2 == 1) { // find out of order positive // element in remaining array swap(arr, i, j); break; } } } } //after modification printArray(arr, n); return 0; } // This code is contributed by AnitAggarwal ```