将所有零移动到数组末尾
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count。 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#
/* 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
// 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)。
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