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推到数组的末尾。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该相同。预期时间复杂度为 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, 0};
Input : arr[] = {1, 2, 0, 0, 0, 3, 6};
Output : arr[] = {1, 2, 3, 6, 0, 0, 0};可以有很多方法来解决这个问题。以下是解决此问题的一种简单而有趣的方法。
从左到右遍历给定数组'arr'。遍历时，保持数组中非零元素的计数。让计数为“计数”。对于每个非零元素 arr[i]，将元素放在“arr[count]”并增加“count”。完成遍历后，所有非零元素都已移动到前端，并且“count”设置为第一个 0 的索引。现在我们需要做的就是运行一个循环，使从“count”到结束的所有元素都为零的数组。
下面是上述方法的实现。
C
// A C program to move all zeroes at the end of array
#include
// 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);
printf("%s\n", "Array after pushing all zeros to end of array:");
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
} C++
#include
#include
#include
void push_zeros_to_end(std::vector& arr)
{
std::stable_partition(arr.begin(),
arr.end(),
[](int n) { return n != 0; });
}
int main()
{
std::vector arr{1,9,8,4,0,0,2,7,0,6,0,9};
push_zeros_to_end(arr);
for(const auto& i : arr)
std::cout << i << ' ';
std::cout << "\n";
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; iPython3
# 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
Javascript
输出
Array after pushing all zeros to end of array :
1 9 8 4 2 7 6 9 0 0 0 0 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)