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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. class Node{
  5. 	public:
  6. 	int data;
  7. 	Node* next;
  8.  
  9. 	Node(int d){
  10. 		data = d;
  11. 		next = NULL;
  12. 	}
  13. };
  14.  
  15. void insert(Node*& head, int data){
  16.  
  17. 	if(head == NULL){
  18. 		head = new Node(data);
  19. 		return;
  20. 	}
  21.  
  22. 	Node* temp = head;
  23. 	while(temp->next != NULL){
  24. 		temp = temp->next;
  25. 	}
  26. 	temp->next = new Node(data);
  27.  
  28. }
  29.  
  30.  
  31. void print(Node* head){
  32. 	while(head!=NULL){
  33. 		cout<<head->data<<" ";
  34. 		head = head->next;
  35. 	}
  36. 	cout<<endl;
  37. }
  38.  
  39. void evenAfterOdd(Node*& head){
  40. 	// To check the approach I took just read the comments at the end of the prog
  41.  
  42. 	Node* evenLast = NULL, *evenFirst = NULL, *oddLast = NULL;
  43.  
  44. 	Node* temp = head;
  45.  
  46. 	while(temp!=NULL){
  47.  
  48. 		if(temp->data%2 == 0){
  49. 			if(evenFirst == NULL){
  50. 				evenFirst = temp;
  51. 				evenLast = temp;
  52. 			}else{
  53. 				evenLast->next = temp;
  54. 				evenLast = temp;
  55. 			}
  56. 		}else{
  57. 			if(oddLast == NULL){
  58. 				head = temp;
  59. 				oddLast = temp;
  60. 			}else{
  61. 				oddLast->next = temp;
  62. 				oddLast = temp;
  63. 			}
  64. 		}
  65. 		temp = temp->next;
  66. 	}
  67.  
  68. 	oddLast->next = evenFirst;
  69. 	evenLast->next = NULL;
  70. }
  71.  
  72. int main() {
  73. 	Node* head = NULL;
  74. 	int n;
  75. 	cin>>n;
  76. 	for(int i = 0; i < n; i++){
  77. 		int temp;
  78. 		cin>>temp;
  79. 		insert(head,temp);
  80. 	}
  81. 	evenAfterOdd(head);
  82. 	print(head);
  83. 	return 0;
  84. }
  85.  
  86.  
  87. 	// Instead of the above approach I will take 2 node pointers
  88. 	// One of them represents the last even node encountered and odd resp
  89. 	// We will also store the first of the Even list as we will do
  90. 	// oddLast->next = evenFirst
Success #stdin #stdout 0s 4380KB
comments (?)
stdin 
5
1 2 2 2 1
stdout 
1 1 2 2 2
https://ideone.com/mwfGY1
Even after odd, Assumption maintain the given order, i.e. We will traverse through the linked list and whenever an even node is encountered remove it and append at the last.
language:
C++14 (gcc 8.3)
created:
6 years ago
visibility:
secret

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