将所有零移动到数组末尾 · GeeksForGeeks 中文系列教程

# 将所有零移动到数组末尾 > 原文： [https://www.geeksforgeeks.org/move-zeroes-end-array/](https://www.geeksforgeeks.org/move-zeroes-end-array/) 给定一个随机数数组，将给定数组的所有零都推到该数组的末尾。例如，如果给定数组为`{1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0}`，则应将其更改为`{1, 9, 8, 4, 2, 7, 6, 0, 0, 0, 0}`。所有其他元素的顺序应相同。预期时间复杂度为`O(n)`，额外空间为`O(1)`。 **示例**： ``` Input : arr[] = {1, 2, 0, 4, 3, 0, 5, 0}; Output : arr[] = {1, 2, 4, 3, 5, 0, 0}; Input : arr[] = {1, 2, 0, 0, 0, 3, 6}; Output : arr[] = {1, 2, 3, 6, 0, 0, 0}; ``` 有很多方法可以解决此问题。以下是解决此问题的简单有趣的方法。从左到右遍历给定的数组“`arr`”。遍历时，维护数组中非零元素的数量。让计数为“计数”。对于每个非零元素`arr[i]`，将其放在“`arr[count]`”处，并递增“`count`”。完全遍历之后，所有非零元素都已移至前端，并且“`count`”被设置为前 0 的索引。现在我们要做的就是运行一个循环，使所有元素从“`count`”到结束都为零。的数组。下面是上述方法的实现。 ## C++ ```cpp // A C++ program to move all zeroes at the end of array #include <iostream> using namespace std; // Function which pushes all zeros to end of an array. void pushZerosToEnd(int arr[], int n) { int count = 0; // Count of non-zero elements // Traverse the array. If element encountered is non- // zero, then replace the element at index 'count' // with this element for (int i = 0; i < n; i++) if (arr[i] != 0) arr[count++] = arr[i]; // here count is // incremented // Now all non-zero elements have been shifted to // front and 'count' is set as index of first 0. // Make all elements 0 from count to end. while (count < n) arr[count++] = 0; } // Driver program to test above function int main() { int arr[] = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}; int n = sizeof(arr) / sizeof(arr[0]); pushZerosToEnd(arr, n); cout << "Array after pushing all zeros to end of array :\n"; for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0; } ``` ## Java ```java /* Java program to push zeroes to back of array */ import java.io.*; class PushZero { // Function which pushes all zeros to end of an array. static void pushZerosToEnd(int arr[], int n) { int count = 0; // Count of non-zero elements // Traverse the array. If element encountered is // non-zero, then replace the element at index 'count' // with this element for (int i = 0; i < n; i++) if (arr[i] != 0) arr[count++] = arr[i]; // here count is // incremented // Now all non-zero elements have been shifted to // front and 'count' is set as index of first 0\. // Make all elements 0 from count to end. while (count < n) arr[count++] = 0; } /*Driver function to check for above functions*/ public static void main (String[] args) { int arr[] = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}; int n = arr.length; pushZerosToEnd(arr, n); System.out.println("Array after pushing zeros to the back: "); for (int i=0; i<n; i++) System.out.print(arr[i]+" "); } } /* This code is contributed by Devesh Agrawal */ ``` ## Python3 ```py # Python3 code to move all zeroes # at the end of array # Function which pushes all # zeros to end of an array. def pushZerosToEnd(arr, n): count = 0 # Count of non-zero elements # Traverse the array. If element # encountered is non-zero, then # replace the element at index # 'count' with this element for i in range(n): if arr[i] != 0: # here count is incremented arr[count] = arr[i] count+=1 # Now all non-zero elements have been # shifted to front and 'count' is set # as index of first 0\. Make all # elements 0 from count to end. while count < n: arr[count] = 0 count += 1 # Driver code arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9] n = len(arr) pushZerosToEnd(arr, n) print("Array after pushing all zeros to end of array:") print(arr) # This code is contributed by "Abhishek Sharma 44" ``` ## C# ```cs /* C# program to push zeroes to back of array */ using System; class PushZero { // Function which pushes all zeros // to end of an array. static void pushZerosToEnd(int []arr, int n) { // Count of non-zero elements int count = 0; // Traverse the array. If element encountered is // non-zero, then replace the element // at index â..countâ.. with this element for (int i = 0; i < n; i++) if (arr[i] != 0) // here count is incremented arr[count++] = arr[i]; // Now all non-zero elements have been shifted to // front and â..countâ.. is set as index of first 0\. // Make all elements 0 from count to end. while (count < n) arr[count++] = 0; } // Driver function public static void Main () { int []arr = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}; int n = arr.Length; pushZerosToEnd(arr, n); Console.WriteLine("Array after pushing all zeros to the back: "); for (int i = 0; i < n; i++) Console.Write(arr[i] +" "); } } /* This code is contributed by Anant Agrawal */ ``` ## PHP ```php <?php // A PHP program to move all // zeroes at the end of array // Function which pushes all // zeros to end of an array. function pushZerosToEnd(&$arr, $n) { // Count of non-zero elements $count = 0; // Traverse the array. If // element encountered is // non-zero, then replace // the element at index // 'count' with this element for ($i = 0; $i < $n; $i++) if ($arr[$i] != 0) // here count is incremented $arr[$count++] = $arr[$i]; // Now all non-zero elements // have been shifted to front // and 'count' is set as index // of first 0\. Make all elements // 0 from count to end. while ($count < $n) $arr[$count++] = 0; } // Driver Code $arr = array(1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9); $n = sizeof($arr); pushZerosToEnd($arr, $n); echo "Array after pushing all " . "zeros to end of array :\n"; for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; // This code is contributed // by ChitraNayal ?> ``` **输出**： ``` Array after pushing all zeros to end of array : 1 9 8 4 2 7 6 9 0 0 0 0 ``` **时间复杂度**：`O(n)`，其中`n`是输入数组中的元素数。 **辅助空间**：`O(1)` 本文由 [**Chandra Prakash**](https://www.facebook.com/chandra.prakash.52643?fref=ts) 贡献。如果发现任何不正确的地方，或者想分享有关上述主题的更多信息，请写评论。