Sorting a linked list is much more complex than sorting an array. Because you need to be wary of links at each step and update them.
The most convenient ways of sorting a linked list is insertion sort, where in you remove one node at a time from list and insert them to the sorted list in correct position.
But merge sorting a linked list uses another approach. It splits the list into two halves, sorts them recursively and then merges them maintaining ascending order of key values.
So the three basic parts of this approach are
You can download the entire program from here
The most convenient ways of sorting a linked list is insertion sort, where in you remove one node at a time from list and insert them to the sorted list in correct position.
But merge sorting a linked list uses another approach. It splits the list into two halves, sorts them recursively and then merges them maintaining ascending order of key values.
So the three basic parts of this approach are
- Splitting the list into two halves of equal length
- Sorting these halves
- Merging the halves
- If the list has more than one node
- Splitting the list into two halves of equal length
- Sorting these halves recursively using steps 2 to 4
- Merging the halves
Split the list into halves
We have seen in this post how to find the mid point of a linked list. Let us use this approach.
We need to use two pointers say slow and fast. Slow moves one node at a time. And fast moves two nodes at a time. When fast has reached end of the list, slow has reached mid point.
And also we should detach first half of the list from second half.
NODEPTR split_list(NODEPTR head) { NODEPTR slow,fast; slow =head; fast = head; if(head==NULL || head->next==NULL) return head; /* if only one node is present, return*/ NODEPTR temp; while(fast!=NULL) { fast = fast->next; if(fast!=NULL){ temp = slow; fast = fast->next; slow = slow->next; } } temp->next = NULL;/*detach first half of list from second half*/ return slow; /*this is the head of second half of list*/ }
Merge sorted sublists
To merge the two sublists which are already sorted we need to compare nodes from two lists, insert the smaller one into the merged list and move to next node in that list. This process has to be repeated until all nodes of both lists are merged. If the two sublists are of unequal length then one list will have nodes remaining, which should be added to merged list
NODEPTR merge_sorted_lists(NODEPTR head1,NODEPTR head2) { NODEPTR newlist = NULL; while(head1!=NULL && head2!=NULL ) { if(head1->val <head2->val) { NODEPTR temp = head1; head1 = head1->next; temp->next = NULL; newlist = append_node(newlist,temp); } else { NODEPTR temp = head2; head2 = head2->next; temp->next = NULL; newlist = append_node(newlist,temp); } } while(head1!=NULL) { NODEPTR temp = head1; head1 = head1->next; temp->next = NULL; newlist = append_node(newlist,temp); } while(head2!=NULL) { NODEPTR temp = head2; head2 = head2->next; temp->next = NULL; newlist = append_node(newlist,temp); } return newlist; }
Merge sort the list
Let us use these two functions in the algorithm given earlier to merge sort the list.
NODEPTR merge_sort(NODEPTR head) { if(head==NULL || head->next==NULL) return head; /* list has one node or is empty*/ NODEPTR mid = split_list(head); head = merge_sort(head);/* sort first sublist*/ mid = merge_sort(mid);/* sort second sublist*/ head = merge_sorted_lists(head,mid);/* merge these*/ return head; }
You can download the entire program from here
Comments
Post a Comment