分离偶数和奇数|套装2

给定大小为N的数组arr [] ，任务是分离偶数和奇数。首先打印所有偶数，然后打印奇数。

例子：

Input: arr[] = {8, 22, 65, 70, 33, 60, 2, 34, 43, 21} 
Output: {8, 22, 70, 60, 2, 34, 65, 33, 43, 21} 

Input: arr[] = {18, 52, 37, 70, 3, 63, 2, 34} 
Output: {18, 52, 70, 2, 34, 37, 3, 63} 

线性方法：此问题可以看作是荷兰国旗问题的一种变体。有关解决问题的所有线性方法，请参阅前一篇文章。

时间复杂度： O(N)
辅助空间： O(1)

使用stable_partition()函数： stable_partition()算法排列由start和end定义的序列，以使由用户定义的谓词函数指定的谓词的所有元素都返回True ，在该函数为其返回False的元素之前。分区是稳定的。因此，保留了序列的相对顺序。

句法：

template 
BiIter stable_partition(BiIter start, BiIter end, UnPred pfn) 

参数：

start: the range of elements to reorder 
end: the range of elements to reorder 
pfn: User-defined predicate function object that defines the condition to be satisfied if an element is to be classified. 
A predicate takes a single argument and returns True or False. 
Return Value: Returns an iterator to the beginning of the elements for which the predicate is false. 

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to segregate
// odd and even numbers
void segregate(vector arr)
{
 
    // Using stable partition
    // with lambda expression
    stable_partition(arr.begin(),
                     arr.end(),
                     [](int a) {
                         return a % 2 == 0;
                     });
 
    // Print array after partition
    for (int num : arr)
        cout << num << " ";
}
 
// Driver Code
int main()
{
    // Given array arr[]
    vector arr = { 18, 52, 37, 70,
                        3, 63, 2, 34 };
 
    // Function Call
    segregate(arr);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{
   
static int[] stable_partition(int arr[])
{
  // Initialize left and
  // right indexes
  int left = 0,
      right = arr.length - 1;
   
  while (left < right)
  {
    // Increment left index while
    // we see 0 at left
    while (arr[left] % 2 == 0 &&
           left < right)
      left++;
 
    // Decrement right index while
    // we see 1 at right
    while (arr[right] % 2 == 1 &&
           left < right)
      right--;
 
    if (left < right)
    {
      // Swap arr[left] and
      // arr[right]
      int temp = arr[left];
      arr[left] = arr[right];
      arr[right] = temp;
      left++;
      right--;
    }
  }
  return arr;
}
 
// Function to segregate
// odd and even numbers
static void segregate(int[] arr)
{
  // Using stable partition
  // with lambda expression
  int []ans = stable_partition(arr);
 
  // Print array after partition
  for (int num : ans)
    System.out.print(num + " ");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given array arr[]
  int[] arr = {18, 52, 37, 70,
               3, 63, 2, 34};
 
  // Function Call
  segregate(arr);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
 
# Function to segregate
# odd and even numbers
def segregate(arr):
     
    # Using stable partition
    # with lambda expression
    odd = []
    even = []
     
    for x in arr:
        if (x % 2 == 0):
            even.append(x)
        else:
            odd.append(x)
             
    for x in even:
        print(x, end = " ")
    for x in odd:
        print(x, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 18, 52, 37, 70, 3, 63, 2, 34 ]
 
    # Function Call
    segregate(arr)
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
   
static void stable_partition(int []arr)
{
  // Initialize left and
  // right indexes
  List odd = new List();
  List even = new List();
 
  foreach(int i in arr)
  {
    if(i % 2 == 1)
      odd.Add(i);
    else
      even.Add(i);
  }
   
  foreach(int i in even)
    Console.Write(i +" ");
   
  foreach(int i in odd)
    Console.Write(i +" ");
}
 
// Function to segregate
// odd and even numbers
static void segregate(int[] arr)
{
  // Using stable partition
  // with lambda expression
  stable_partition(arr);
 
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array []arr
  int[] arr = {18, 52, 37, 70,
               3, 63, 2, 34};
 
  // Function Call
  segregate(arr);
}
}
 
// This code is contributed by 29AjayKumar

输出

18 52 70 2 34 37 3 63 

时间复杂度： O(N)
辅助空间： O(1)