Thursday, September 7th

 Reverse part of a linked list

Given a linked list and 2 positions left and right, reverse the linked list in between the 2 positions

The football play portrayed above is an end around, sometimes a precursor to a reverse

I came up with a couple of solutions that solved the problem by swapping values between nodes the being reversed, but as some problem commenters pointed out, this isn't exactly the same as reversing part of the list.  

Moreover, one could easily imagine being asked this question in an interview, and upon further clarification, begin required to reverse the whole list, not just values.  

Reversing the actual list nodes takes a little more thought, so we'll present that here.

We start the function by checking for an empty head

ListNode* reverseBetween(ListNode* head, int left, int right) {
if(head==NULL) return NULL;
        ...
    }

Next we declare 2 pointers, current and previous.  Then we iterate over the list left-1 times, updating the previous pointer to current, and the current to the next node.  The reason we iterate only left-1 times is because the list is indexed at 1

ListNode *curr=head;
ListNode *prev=NULL;

for(int i=0; i<left-1; i++) {
prev=curr;
curr=curr->next;
}

Before moving forward, we declare 2 pointers that will help connect the list together once the part of it is reversed.  First is a pointer that will point from the beginning part of the list to the start of the reversed part.

ListNode* list_to_reversal_start=prev;

We also declare a pointer that will connect the end of the reversed part of the list to the proper end of the list

ListNode* reversal_end_to_list=curr;

Now we iterate over the number of nodes we need to reverse.  First we declare a third pointer nextnode which points to the next node after current.  Then within the for loop, we set next node equal to current next, set current next to the previous node (reversing it), then set previous to current, and current to the next node.  We are using 3 pointers to move forward in the linked list while reversing it.

ListNode* nextnode=NULL;
for(int i=0; i<=right-left; i++) {
nextnode=curr->next;
curr->next=prev;
prev=curr;
curr=nextnode;
}

Once the part of the list is reversed, we connect it back with the original list.

If the start of the list pointer isn't null, we point the next node to the end of the reversed list.  If it is null, it means that the reversal asked to head node, so set head equal to the end of the reversed list. 

if(list_to_reversal_start!=NULL) {
list_to_reversal_start->next=prev;
} else {
head=prev;
}

Finally, set the end of the reversed list to the current node, the start of the unsorted list's end

reversal_end_to_list->next=curr;

Return head

return head;

In an interview it would be important to ask if swapping values is acceptable.  But prepare for the worst case.  This problem demonstrates how to use 3 pointers for moving forward while reversing the list.

Thanks to the Editorial for demonstrating 2 interesting solutions

Comments

Post a Comment

Popular posts from this blog

Wednesday, September 20th

Image
 Reduce X to Zero Given an integer array and value x, use elements at the right and left of the array to reduce the x value to zero A collection of windows, by Daderot .  Shared under the Creative Commons Attribution-Share Alike 3.0 Unported License .  No changes made A key observation is that if elements at the ends of the array are used to sum up to a target value x , then elements in a contiguous subarray will equal the total_sum - x .  So we will look for the largest window equal to  total_sum - x Since all elements in the array are positive, then we can solve the problem using 2 pointers that form a sliding window. The general idea is to begin 2 pointers right and left at index 0.  Continue incrementing the right index by 1, adding up the elements it passes into the current_window  sum.  If the sum equals the target value, update the length.  Otherwise, if the sum is greater than the target sum, increment the left index until the window...
Read more

Monday, September 11th

Image
 Groups There are a given number of people, each of whom is assigned to a group.  The number identifier of the group is also its size.  The size of a group may be a multiple of its size, e.g. a group of 3 may have 6 people.  Return the  collection of all groups . Team USA Softball (a group).  Shared by USASoftball , via Wikipedia , distributed under the Creative Commons Attribution-Share Alike 3.0 Unported license.  No changes made. This problem is pretty straightforward.  The idea is to create a map (or a vector) keyed by the group, storing an array of individuals belonging to the group. First we create the vector-of-vectors to store our solution, then we initialize a vector of vectors that will store individuals according to their group numbers.  The size of the solution vector is n+1, because contrary to what the beginning of the problem statement said, the indices/group sizes go from 1 to n, according to the stated constraints. vect...
Read more

Tuesday, September 19th

Image
 Find the duplicate number Given an array of integers of length n+1, which contains numbers within the range [1,n], and contains on element that is duplicated at least once, find the duplicate element , considering use of only constant space and no changes to the array "How They Met Themselves" by Dante Rossetti, depicting doppelgängers The problem of finding duplicates is made difficult by the constraints of no modifications to the original array, and use of only constant extra space.  Still, despite these constraints there are still several ways to solve the problem.   Since yesterday's solution involved binary search, we will hone our BS skills by applying it to today's problem The general idea is based on the observation that without duplicates, the number of elements less than or equal an element is equal to the element itself.  E.g., in the collection {1,2,3,4}, the number of elements less than or equal to 2 is 2, less than or equal to 3 is 3, etc.   Howe...
Read more