给定一个由N 个正整数组成的数组arr[] ,具有相等数量的偶数和奇数元素。任务是使用就地交换来交换数组中偶数和奇数元素的位置。
例子:
Input: arr[] = {1, 3, 2, 4}
Output: 2 4 1 3
Explanation:
Before rearranging the given array, indices 0 and 1 had odd elements and indices 2 and 3 had even elements.
After rearrangement, array becomes {2, 4, 1, 3} where indices 0 and 1 have even elements and indices 2 and 3 have odd elements.
Input: arr[] = {2, 2, 1, 3}
Output: 1 3 2 2
Explanation:
Before rearranging the given array, indices 0 and 1 had even elements and indices 2 and 3 had odd elements.
After rearrangement, array becomes {1, 3, 2, 2} where indices 0 and 1 have odd elements and indices 2 and 3 have even elements.
朴素的方法:最简单的方法是使用两个循环遍历数组元素,外循环选取数组的每个元素,内循环为选取的元素找到相反的奇偶校验元素并交换它们。交换元素后,将选取的元素标记为负值,以免再次选取。最后,使所有元素为正并打印数组。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to replace each even
// element by odd and vice-versa
// in a given array
void replace(int arr[], int n)
{
// Traverse array
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
// If current element is even
// then swap it with odd
if (arr[i] >= 0 && arr[j] >= 0 &&
arr[i] % 2 == 0 &&
arr[j] % 2 != 0)
{
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break;
}
// If current element is odd
// then swap it with even
else if (arr[i] >= 0 && arr[j] >= 0 &&
arr[i] % 2 != 0 &&
arr[j] % 2 == 0)
{
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break;
}
}
}
// Marked element positive
for(int i = 0; i < n; i++)
arr[i] = abs(arr[i]);
// Print final array
for(int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 3, 2, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
replace(arr,n);
}
// This code is contributed by SURENDRA_GANGWAR
Java
// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG {
// Function to replace each even
// element by odd and vice-versa
// in a given array
static void replace(int[] arr)
{
// Stores length of array
int n = arr.length;
// Traverse array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// If current element is even
// then swap it with odd
if (arr[i] >= 0
&& arr[j] >= 0
&& arr[i] % 2 == 0
&& arr[j] % 2 != 0) {
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break;
}
// If current element is odd
// then swap it with even
else if (arr[i] >= 0
&& arr[j] >= 0
&& arr[i] % 2 != 0
&& arr[j] % 2 == 0) {
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break;
}
}
}
// Marked element positive
for (int i = 0; i < n; i++)
arr[i] = Math.abs(arr[i]);
// Print final array
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int[] arr = { 1, 3, 2, 4 };
// Function Call
replace(arr);
}
}
Python3
# Python3 program for the
# above approach
# Function to replace each even
# element by odd and vice-versa
# in a given array
def replace(arr, n):
# Traverse array
for i in range(n):
for j in range(i + 1, n):
# If current element is
# even then swap it with odd
if (arr[i] >= 0 and
arr[j] >= 0 and
arr[i] % 2 == 0 and
arr[j] % 2 != 0):
# Perform Swap
tmp = arr[i]
arr[i] = arr[j]
arr[j] = tmp
# Change the sign
arr[j] = -arr[j]
break
# If current element is odd
# then swap it with even
elif (arr[i] >= 0 and
arr[j] >= 0 and
arr[i] % 2 != 0 and
arr[j] % 2 == 0):
# Perform Swap
tmp = arr[i]
arr[i] = arr[j]
arr[j] = tmp
# Change the sign
arr[j] = -arr[j]
break
# Marked element positive
for i in range(n):
arr[i] = abs(arr[i])
# Print final array
for i in range(n):
print(arr[i], end = " ")
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [1, 3, 2, 4]
n = len(arr)
# Function Call
replace(arr, n)
# This code is contributed by Chitranayal
C#
// C# program for the above approach
using System;
class GFG{
// Function to replace each even
// element by odd and vice-versa
// in a given array
static void replace(int[] arr)
{
// Stores length of array
int n = arr.Length;
// Traverse array
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
// If current element is even
// then swap it with odd
if (arr[i] >= 0 &&
arr[j] >= 0 &&
arr[i] % 2 == 0 &&
arr[j] % 2 != 0)
{
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break;
}
// If current element is odd
// then swap it with even
else if (arr[i] >= 0 &&
arr[j] >= 0 &&
arr[i] % 2 != 0 &&
arr[j] % 2 == 0)
{
// Perform Swap
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
// Change the sign
arr[j] = -arr[j];
break;
}
}
}
// Marked element positive
for(int i = 0; i < n; i++)
arr[i] = Math.Abs(arr[i]);
// Print readonly array
for(int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int[] arr = { 1, 3, 2, 4 };
// Function Call
replace(arr);
}
}
// This code is contributed by Rajput-Ji
Javascript
C++
// C++ program for the above approach
#include
using namespace std;
#define N 3
#define M 4
// Function to replace odd elements
// with even elements and vice versa
void swapEvenOdd(int arr[], int n)
{
int o = -1, e = -1;
// Traverse the given array
for(int i = 0; i < n; i++)
{
// If arr[i] is visited
if (arr[i] < 0)
continue;
int r = -1;
if (arr[i] % 2 == 0)
{
o++;
// Find the next odd element
while (arr[o] % 2 == 0 || arr[o] < 0)
o++;
r = o;
}
else
{
e++;
// Find next even element
while (arr[e] % 2 == 1 || arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the final array
for(int i = 0; i < n; i++)
{
cout << (-1 * arr[i]) << " ";
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 3, 2, 4 };
// Length of the array
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
swapEvenOdd(arr, n);
}
// This code is contributed by Rajput-Ji
Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd(int arr[])
{
// Length of the array
int n = arr.length;
int o = -1, e = -1;
// Traverse the given array
for (int i = 0; i < n; i++) {
// If arr[i] is visited
if (arr[i] < 0)
continue;
int r = -1;
if (arr[i] % 2 == 0) {
o++;
// Find the next odd element
while (arr[o] % 2 == 0
|| arr[o] < 0)
o++;
r = o;
}
else {
e++;
// Find next even element
while (arr[e] % 2 == 1
|| arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the final array
for (int i = 0; i < n; i++) {
System.out.print(
(-1 * arr[i]) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 1, 3, 2, 4 };
// Function Call
swapEvenOdd(arr);
}
}
Python3
# Python3 program for the above approach
# Function to replace odd elements
# with even elements and vice versa
def swapEvenOdd(arr):
# Length of the array
n = len(arr)
o = -1
e = -1
# Traverse the given array
for i in range(n):
# If arr[i] is visited
if (arr[i] < 0):
continue
r = -1
if (arr[i] % 2 == 0):
o += 1
# Find the next odd element
while (arr[o] % 2 == 0 or
arr[o] < 0):
o += 1
r = o
else:
e += 1
# Find next even element
while (arr[e] % 2 == 1 or
arr[e] < 0):
e += 1
r = e
# Mark them visited
arr[i] *= -1
arr[r] *= -1
# Swap them
tmp = arr[i]
arr[i] = arr[r]
arr[r] = tmp
# Print the final array
for i in range(n):
print((-1 * arr[i]), end = " ")
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [ 1, 3, 2, 4 ]
# Function Call
swapEvenOdd(arr)
# This code is contributed by Amit Katiyar
C#
// C# program for the above approach
using System;
class GFG{
// Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd(int []arr)
{
// Length of the array
int n = arr.Length;
int o = -1, e = -1;
// Traverse the given array
for(int i = 0; i < n; i++)
{
// If arr[i] is visited
if (arr[i] < 0)
continue;
int r = -1;
if (arr[i] % 2 == 0)
{
o++;
// Find the next odd element
while (arr[o] % 2 == 0 ||
arr[o] < 0)
o++;
r = o;
}
else
{
e++;
// Find next even element
while (arr[e] % 2 == 1 ||
arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the readonly array
for(int i = 0; i < n; i++)
{
Console.Write((-1 * arr[i]) + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 1, 3, 2, 4 };
// Function Call
swapEvenOdd(arr);
}
}
// This code is contributed by Amit Katiyar
Javascript
C++
// C++ program for the above approach
#include
using namespace std;
// Function to replace odd elements
// with even elements and vice versa
void swapEvenOdd(int arr[], int N)
{
stack > stack;
// Push the first element to stack
stack.push({ 0, arr[0] });
// iterate the array and swap even and odd
for (int i = 1; i < N; i++)
{
if (!stack.empty())
{
if (arr[i] % 2 != stack.top().second % 2)
{
// pop and swap
pair pop = stack.top();
stack.pop();
int index = pop.first, val = pop.second;
arr[index] = arr[i];
arr[i] = val;
}
else
stack.push({ i, arr[i] });
}
else
stack.push({ i, arr[i] });
}
// print the arr[]
for (int i = 0; i < N; i++)
cout << arr[i] << " ";
}
// Driven Program
int main()
{
// Given array arr[]
int arr[] = { 1, 3, 2, 4 };
// Stores the length of array
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
swapEvenOdd(arr, N);
return 0;
}
// This code is contributed by Kingash.
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd(int arr[], int N)
{
Stack stack = new Stack<>();
// Push the first element to stack
stack.push(new int[] { 0, arr[0] });
// iterate the array and swap even and odd
for (int i = 1; i < N; i++)
{
if (!stack.isEmpty())
{
if (arr[i] % 2 != stack.peek()[1] % 2)
{
// pop and swap
int pop[] = stack.pop();
int index = pop[0], val = pop[1];
arr[index] = arr[i];
arr[i] = val;
}
else
stack.push(new int[] { i, arr[i] });
}
else
stack.push(new int[] { i, arr[i] });
}
// print the arr[]
for (int i = 0; i < N; i++)
System.out.print(arr[i] + " ");
}
// Driver code
public static void main(String[] args)
{
// Given array arr
int arr[] = { 1, 3, 2, 4 };
// length of the arr
int N = arr.length;
// Function Call
swapEvenOdd(arr, N);
}
}
// This code is contributed by Kingash.
Python3
# Python3 program for the above approach
# Function to replace odd elements
# with even elements and vice versa
def swapEvenOdd(arr):
stack = []
# Push the first element to stack
stack.append((0, arr[0],))
# iterate the array and swap even and odd
for i in range(1, len(arr)):
if stack:
if arr[i]%2 != stack[-1][1]%2:
#pop and swap
index, val = stack.pop(-1)
arr[index] = arr[i]
arr[i] = val
else:
stack.append((i, arr[i],))
else:
stack.append((i, arr[i],))
return arr
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [ 1, 3, 2, 4 ]
# Function Call
print(swapEvenOdd(arr))
# This code is contributed by Arunabha Choudhury
2 4 1 3
时间复杂度: O(N 2 )
辅助空间: O(N)
高效的方法:为了优化上述方法,思想是使用两个指针方法。通过拾取大于0的每个元素遍历数组并搜索大于0从当前索引到该阵列的端部的相反奇偶性元件更大。如果找到,交换元素并将它们与-1相乘。请按照以下步骤解决问题:
- 用-1初始化变量e和o ,这将分别存储当前找到的尚未被采用的偶数和奇数。
- 在[0, N – 1]范围内遍历给定数组并执行以下操作:
- 如果元素arr[i]大于0 ,则选择它。
- 如果arr[i]是偶数,则将o + 1增加1并找到下一个尚未标记的奇数。通过将当前和找到的数字乘以-1并交换它们来标记它们。
- 如果arr[i]是奇数,则将e + 1增加1并找到下一个尚未标记的偶数。通过将当前和找到的数字乘以-1并交换它们来标记它们。
- 遍历整个数组后,将其元素乘以-1再次使它们为正数并打印该数组。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
#define N 3
#define M 4
// Function to replace odd elements
// with even elements and vice versa
void swapEvenOdd(int arr[], int n)
{
int o = -1, e = -1;
// Traverse the given array
for(int i = 0; i < n; i++)
{
// If arr[i] is visited
if (arr[i] < 0)
continue;
int r = -1;
if (arr[i] % 2 == 0)
{
o++;
// Find the next odd element
while (arr[o] % 2 == 0 || arr[o] < 0)
o++;
r = o;
}
else
{
e++;
// Find next even element
while (arr[e] % 2 == 1 || arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the final array
for(int i = 0; i < n; i++)
{
cout << (-1 * arr[i]) << " ";
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 3, 2, 4 };
// Length of the array
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
swapEvenOdd(arr, n);
}
// This code is contributed by Rajput-Ji
Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd(int arr[])
{
// Length of the array
int n = arr.length;
int o = -1, e = -1;
// Traverse the given array
for (int i = 0; i < n; i++) {
// If arr[i] is visited
if (arr[i] < 0)
continue;
int r = -1;
if (arr[i] % 2 == 0) {
o++;
// Find the next odd element
while (arr[o] % 2 == 0
|| arr[o] < 0)
o++;
r = o;
}
else {
e++;
// Find next even element
while (arr[e] % 2 == 1
|| arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the final array
for (int i = 0; i < n; i++) {
System.out.print(
(-1 * arr[i]) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 1, 3, 2, 4 };
// Function Call
swapEvenOdd(arr);
}
}
蟒蛇3
# Python3 program for the above approach
# Function to replace odd elements
# with even elements and vice versa
def swapEvenOdd(arr):
# Length of the array
n = len(arr)
o = -1
e = -1
# Traverse the given array
for i in range(n):
# If arr[i] is visited
if (arr[i] < 0):
continue
r = -1
if (arr[i] % 2 == 0):
o += 1
# Find the next odd element
while (arr[o] % 2 == 0 or
arr[o] < 0):
o += 1
r = o
else:
e += 1
# Find next even element
while (arr[e] % 2 == 1 or
arr[e] < 0):
e += 1
r = e
# Mark them visited
arr[i] *= -1
arr[r] *= -1
# Swap them
tmp = arr[i]
arr[i] = arr[r]
arr[r] = tmp
# Print the final array
for i in range(n):
print((-1 * arr[i]), end = " ")
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [ 1, 3, 2, 4 ]
# Function Call
swapEvenOdd(arr)
# This code is contributed by Amit Katiyar
C#
// C# program for the above approach
using System;
class GFG{
// Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd(int []arr)
{
// Length of the array
int n = arr.Length;
int o = -1, e = -1;
// Traverse the given array
for(int i = 0; i < n; i++)
{
// If arr[i] is visited
if (arr[i] < 0)
continue;
int r = -1;
if (arr[i] % 2 == 0)
{
o++;
// Find the next odd element
while (arr[o] % 2 == 0 ||
arr[o] < 0)
o++;
r = o;
}
else
{
e++;
// Find next even element
while (arr[e] % 2 == 1 ||
arr[e] < 0)
e++;
r = e;
}
// Mark them visited
arr[i] *= -1;
arr[r] *= -1;
// Swap them
int tmp = arr[i];
arr[i] = arr[r];
arr[r] = tmp;
}
// Print the readonly array
for(int i = 0; i < n; i++)
{
Console.Write((-1 * arr[i]) + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 1, 3, 2, 4 };
// Function Call
swapEvenOdd(arr);
}
}
// This code is contributed by Amit Katiyar
Javascript
2 4 1 3
时间复杂度: O(N)
辅助空间: O(N)
有效方法2:
另一种方法是使用堆栈。请按照以下步骤操作:
- 从数组 arr[] 中取出索引处的元素并压入堆栈
- 将 arr[] 从索引 i = 1 迭代到 end 并执行以下操作:
- 如果堆栈顶部的 arr[i] 和 item 不是偶数或不是奇数,则弹出和交换
- 否则将项目推入堆栈
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to replace odd elements
// with even elements and vice versa
void swapEvenOdd(int arr[], int N)
{
stack > stack;
// Push the first element to stack
stack.push({ 0, arr[0] });
// iterate the array and swap even and odd
for (int i = 1; i < N; i++)
{
if (!stack.empty())
{
if (arr[i] % 2 != stack.top().second % 2)
{
// pop and swap
pair pop = stack.top();
stack.pop();
int index = pop.first, val = pop.second;
arr[index] = arr[i];
arr[i] = val;
}
else
stack.push({ i, arr[i] });
}
else
stack.push({ i, arr[i] });
}
// print the arr[]
for (int i = 0; i < N; i++)
cout << arr[i] << " ";
}
// Driven Program
int main()
{
// Given array arr[]
int arr[] = { 1, 3, 2, 4 };
// Stores the length of array
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
swapEvenOdd(arr, N);
return 0;
}
// This code is contributed by Kingash.
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Function to replace odd elements
// with even elements and vice versa
static void swapEvenOdd(int arr[], int N)
{
Stack stack = new Stack<>();
// Push the first element to stack
stack.push(new int[] { 0, arr[0] });
// iterate the array and swap even and odd
for (int i = 1; i < N; i++)
{
if (!stack.isEmpty())
{
if (arr[i] % 2 != stack.peek()[1] % 2)
{
// pop and swap
int pop[] = stack.pop();
int index = pop[0], val = pop[1];
arr[index] = arr[i];
arr[i] = val;
}
else
stack.push(new int[] { i, arr[i] });
}
else
stack.push(new int[] { i, arr[i] });
}
// print the arr[]
for (int i = 0; i < N; i++)
System.out.print(arr[i] + " ");
}
// Driver code
public static void main(String[] args)
{
// Given array arr
int arr[] = { 1, 3, 2, 4 };
// length of the arr
int N = arr.length;
// Function Call
swapEvenOdd(arr, N);
}
}
// This code is contributed by Kingash.
蟒蛇3
# Python3 program for the above approach
# Function to replace odd elements
# with even elements and vice versa
def swapEvenOdd(arr):
stack = []
# Push the first element to stack
stack.append((0, arr[0],))
# iterate the array and swap even and odd
for i in range(1, len(arr)):
if stack:
if arr[i]%2 != stack[-1][1]%2:
#pop and swap
index, val = stack.pop(-1)
arr[index] = arr[i]
arr[i] = val
else:
stack.append((i, arr[i],))
else:
stack.append((i, arr[i],))
return arr
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [ 1, 3, 2, 4 ]
# Function Call
print(swapEvenOdd(arr))
# This code is contributed by Arunabha Choudhury
[4, 2, 3, 1]
时间复杂度: O(N)
辅助空间: O(N)
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