# 将所有零移动到数组的末尾| 系列 2（使用单遍历） > 原文： [https://www.geeksforgeeks.org/move-zeroes-end-array-set-2-using-single-traversal/](https://www.geeksforgeeks.org/move-zeroes-end-array-set-2-using-single-traversal/) 给定 **n** 个数字的数组。 问题是将所有 0 移到数组的末尾，同时保持其他元素的顺序。 只需要数组的一次遍历。 例子： ``` Input : arr[] = {1, 2, 0, 0, 0, 3, 6} Output : 1 2 3 6 0 0 0 Input: arr[] = {0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9} Output: 1 9 8 4 2 7 6 9 0 0 0 0 0 ``` **算法**： ``` moveZerosToEnd(arr, n) Initialize count = 0 for i = 0 to n-1 if (arr[i] != 0) then swap(arr[count++], arr[i]) ``` ## CPP ``` // C++ implementation to move all zeroes at // the end of array #include <iostream> using namespace std; // function to move all zeroes at // the end of array void moveZerosToEnd(int arr[], int n) {     // Count of non-zero elements     int count = 0;     // Traverse the array. If arr[i] is non-zero, then     // swap the element at index 'count' with the     // element at index 'i'     for (int i = 0; i < n; i++)         if (arr[i] != 0)             swap(arr[count++], arr[i]); } // function to print the array elements void printArray(int arr[], int n) {     for (int i = 0; i < n; i++)         cout << arr[i] << " "; } // Driver program to test above int main() {     int arr[] = { 0, 1, 9, 8, 4, 0, 0, 2,                          7, 0, 6, 0, 9 };     int n = sizeof(arr) / sizeof(arr[0]);     cout << "Original array: ";     printArray(arr, n);     moveZerosToEnd(arr, n);     cout << "\nModified array: ";     printArray(arr, n);     return 0; } ``` ## Java ```java // Java implementation to move  // all zeroes at the end of array import java.io.*; class GFG { // function to move all zeroes at // the end of array static void moveZerosToEnd(int arr[], int n) {     // Count of non-zero elements     int count = 0;     int temp;     // Traverse the array. If arr[i] is      // non-zero, then swap the element at      // index 'count' with the element at      // index 'i'     for (int i = 0; i < n; i++) {     if ((arr[i] != 0)) {         temp = arr[count];         arr[count] = arr[i];         arr[i] = temp;         count = count + 1;     }     } } // function to print the array elements static void printArray(int arr[], int n) {     for (int i = 0; i < n; i++)     System.out.print(arr[i] + " "); } // Driver program to test above public static void main(String args[]) {     int arr[] = {0, 1, 9, 8, 4, 0, 0, 2,                          7, 0, 6, 0, 9};     int n = arr.length;     System.out.print("Original array: ");     printArray(arr, n);     moveZerosToEnd(arr, n);     System.out.print("\nModified array: ");     printArray(arr, n); } } // This code is contributed by Nikita Tiwari. ``` ## Python3 ```py # Python implementation to move all zeroes at # the end of array # function to move all zeroes at # the end of array def moveZerosToEnd (arr, n):     # Count of non-zero elements     count = 0;     # Traverse the array. If arr[i] is non-zero, then     # swap the element at index 'count' with the     # element at index 'i'     for i in range(0, n):         if (arr[i] != 0):             arr[count], arr[i] = arr[i], arr[count]             count+=1 # function to print the array elements def printArray(arr, n):     for i in range(0, n):         print(arr[i],end=" ") # Driver program to test above arr = [ 0, 1, 9, 8, 4, 0, 0, 2,     7, 0, 6, 0, 9 ] n = len(arr) print("Original array:", end=" ") printArray(arr, n) moveZerosToEnd(arr, n) print("\nModified array: ", end=" ") printArray(arr, n) # This code is contributed by # Smitha Dinesh Semwal ``` ## C# ```cs // C# implementation to move // all zeroes at the end of array using System; class GFG {     // function to move all zeroes at     // the end of array     static void moveZerosToEnd(int[] arr, int n)     {         // Count of non-zero elements         int count = 0;         int temp;         // Traverse the array. If arr[i] is         // non-zero, then swap the element at         // index 'count' with the element at         // index 'i'         for (int i = 0; i < n; i++)         {             if ((arr[i] != 0))              {                 temp = arr[count];                 arr[count] = arr[i];                 arr[i] = temp;                 count = count + 1;             }         }     }     // function to print the array elements     static void printArray(int[] arr, int n)     {         for (int i = 0; i < n; i++)             Console.Write(arr[i] + " ");     }     // Driver program to test above     public static void Main()     {         int[] arr = { 0, 1, 9, 8, 4, 0, 0, 2,                     7, 0, 6, 0, 9 };         int n = arr.Length;         Console.Write("Original array: ");         printArray(arr, n);         moveZerosToEnd(arr, n);         Console.Write("\nModified array: ");         printArray(arr, n);     } } // This code is contributed by Sam007 ``` Output: ``` Original array: 0 1 9 8 4 0 0 2 7 0 6 0 9 Modified array: 1 9 8 4 2 7 6 9 0 0 0 0 0 ``` 时间复杂度：`O(n)`。 辅助空间：`O(1)`。 * * * * * *