算法设计（将所有零移动到数组末尾详细代码实现） _数据结构

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

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, 如果给定的数组是{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。预期的时间复杂度为O(n), 额外空间为O(1)。
例子：

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};

推荐：请在"实践首先, 在继续解决方案之前。
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 ++

// 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 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

# 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# 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 a..counta.. 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 a..counta.. 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 */

的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 ?>

输出如下：