Problem Statement
The task at hand is to sort a linked list, which is a data structure consisting of nodes where each node contains a value and a reference (or link) to the next node in the sequence. Unlike arrays, linked lists do not have indices for direct access, making some operations, like sorting, more complex. The function needs to accept the head of the linked list (i.e., the first node in the list) and return it sorted in ascending order based on the values in the nodes.
Examples
Example 1
Input:
head = [4,2,1,3]
Output:
[1,2,3,4]
Example 2
Input:
head = [-1,5,3,4,0]
Output:
[-1,0,3,4,5]
Example 3
Input:
head = []
Output:
[]
Constraints
- The number of nodes in the list is in the range
[0, 5 * 104]. -105 <= Node.val <= 105
Approach and Intuition
When approaching the problem of sorting a linked list, several considerations and strategies can be taken based on the given constraints and the nature of linked lists:
Sorting Strategy:
- Choosing the right sorting algorithm is critical. Given the nature of linked lists (i.e., nodes connected via pointers), some algorithms that require random access, like quicksort, are less efficient than those that can efficiently handle sequences, like merge sort or insertion sort.
Implementation Steps:
If the linked list is empty or contains a single element, it is already sorted. Return the linked list as is.
For the merge sort approach:
- Split the linked list into two halves. This can be efficiently done using the fast and slow pointer technique, where the slow pointer advances one step for every two steps the fast pointer takes. When the fast pointer reaches the end, the slow pointer will be at the midpoint.
- Recursively apply merge sort to both halves.
- Merge the two sorted halves. This step involves re-linking nodes such that they preserve the sorted order.
For insertion sort:
- Create a new linked list to act as the sorted portion.
- Traverse each node in the original list, locate its proper position in the sorted list, and re-link the nodes accordingly.
Considerations:
- Time Complexity: Sorting algorithms like merge sort have a better worst-case time complexity (O(n log n)) for linked lists due to the linked nature of the data.
- Space Complexity: Depending on the sorting algorithm, extra space may be required for the sorting process. For example, merge sort on linked lists can be implemented in O(1) extra space if nodes are re-used instead of creating new ones.
By understanding these steps and considerations, the sorting of a linked list can be implemented effectively while managing the complexities related to its non-continuous storage in memory.
Solutions
- C++
- Java
- C
- JavaScript
- Python
class Solution {
public:
ListNode* endNode = new ListNode();
ListNode* nextPart = new ListNode();
ListNode* sortList(ListNode* head) {
if (!head || !head->next) return head;
int nodeCount = getCount(head);
ListNode* initiate = head;
ListNode dummyHead(0);
for (int step = 1; step < nodeCount; step *= 2) {
endNode = &dummyHead;
while (initiate) {
if (!initiate->next) {
endNode->next = initiate;
break;
}
ListNode* midpoint = split(initiate, step);
merge(initiate, midpoint);
initiate = nextPart;
}
initiate = dummyHead.next;
}
return dummyHead.next;
}
ListNode* split(ListNode* begin, int stepLength) {
ListNode* midPrevious = begin;
ListNode* tailPosition = begin->next;
for (int idx = 1; idx < stepLength && (midPrevious->next || tailPosition->next);
idx++) {
if (tailPosition->next) {
tailPosition = (tailPosition->next->next) ? tailPosition->next->next : tailPosition->next;
}
if (midPrevious->next) {
midPrevious = midPrevious->next;
}
}
ListNode* middle = midPrevious->next;
nextPart = tailPosition->next;
midPrevious->next = nullptr;
tailPosition->next = nullptr;
return middle;
}
void merge(ListNode* first, ListNode* second) {
ListNode dummyHead(0);
ListNode* currentTail = &dummyHead;
while (first && second) {
if (first->val < second->val) {
currentTail->next = first;
first = first->next;
currentTail = currentTail->next;
} else {
currentTail->next = second;
second = second->next;
currentTail = currentTail->next;
}
}
currentTail->next = (first) ? first : second;
while (currentTail->next) {
currentTail = currentTail->next;
}
endNode->next = dummyHead.next;
endNode = currentTail;
}
int getCount(ListNode* node) {
int total = 0;
while (node) {
node = node->next;
total++;
}
return total;
}
};
This summary explains a solution for sorting a linked list using an efficient sorting algorithm implemented in C++. The provided code defines a class, Solution, which contains methods to sort the linked list, to split the list into parts, and to merge them, along with a method to count the nodes in the list.
- Begin by checking if the head of the list is null or if there is only one node, in which case the head is returned directly as the sorted list.
- Calculate the total node count using the
getCountfunction to determine the number of steps needed for the sorting process. - Utilize a bottom-up merge sort approach, which merges pairs of small list segments into larger, ordered segments.
- Use two helper functions,
splitandmerge, where:splitdivides the list into two parts at specified intervals and returns the head of the second segment.mergecombines two sorted linked list segments into a single sorted segment.
- This procedure is repeated with a progressively increased step size (
step *= 2)—each cycle takes segments of double the size from the previous iteration and merges them. - Keep tracking the end of the sorted portion of the list and the start of the next part to be processed by maintaining pointers.
- On completion of merging steps for all segment sizes, the head of the newly sorted list, maintained by
dummyHead.next, is returned as the result.
The use of a bottom-up merge sort mechanism allows for sorting in O(n log n) time complexity, which is optimal for linked list sorting operations. No extra space for array conversion is needed, and the approach handles the linked list nodes directly, making it space-efficient barring recursive stack space, which is minimal in this non-recursive implementation.
class Solution {
ListNode endNode = new ListNode();
ListNode nextPartition = new ListNode();
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) return head;
int totalNodes = getCount(head);
ListNode currentHead = head;
ListNode tempHead = new ListNode();
for (int size = 1; size < totalNodes; size = size * 2) {
endNode = tempHead;
while (currentHead != null) {
if (currentHead.next == null) {
endNode.next = currentHead;
break;
}
ListNode midPoint = divide(currentHead, size);
mergeLists(currentHead, midPoint);
currentHead = nextPartition;
}
currentHead = tempHead.next;
}
return tempHead.next;
}
ListNode divide(ListNode start, int size) {
ListNode midPointPrev = start;
ListNode end = start.next;
for (
int index = 1;
index < size && (midPointPrev.next != null || end.next != null);
index++
) {
if (end.next != null) {
end = (end.next.next != null) ? end.next.next : end.next;
}
if (midPointPrev.next != null) {
midPointPrev = midPointPrev.next;
}
}
ListNode midPoint = midPointPrev.next;
midPointPrev.next = null;
nextPartition = end.next;
end.next = null;
return midPoint;
}
void mergeLists(ListNode list1, ListNode list2) {
ListNode dummyNode = new ListNode();
ListNode currentTail = dummyNode;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
currentTail.next = list1;
list1 = list1.next;
currentTail = currentTail.next;
} else {
currentTail.next = list2;
list2 = list2.next;
currentTail = currentTail.next;
}
}
currentTail.next = (list1 != null) ? list1 : list2;
while (currentTail.next != null) {
currentTail = currentTail.next;
}
endNode.next = dummyNode.next;
endNode = currentTail;
}
int getCount(ListNode head) {
int count = 0;
ListNode current = head;
while (current != null) {
current = current.next;
count++;
}
return count;
}
}
The given Java code implements a sorting algorithm for a linked list using a strategy similar to merge sort. Here is a breakdown of how it processes:
sortListfunction:- Handles the case where the list is empty or has only one node.
- Counts the total number of nodes.
- Initiates sorting by breaking down the list into sublists of increasing sizes until the entire list is sorted.
dividefunction:- Splits the list into two parts based on the specified size.
- Returns the starting node of the second part while adjusting pointers to detach these sublists correctly.
mergeListsfunction:- Merges two sorted sublists using a dummy node to facilitate the linking process.
- This function ensures merged output remains sorted.
getCountfunction:- Simply counts the total nodes in the list to help with the divide process.
Overall, this implementation efficiently sorts the linked list in O(n log n) time complexity, typical of merge sort algorithms, but does so non-recursively with linked list nodes to handle large data sizes more effectively. Following these steps, you achieve a sorted version of the input linked list in an optimized manner, leveraging the robust structure and capabilities of Java.
typedef struct ListNode ListNode;
// Function to compute the total nodes present.
int getNodeCount(ListNode *head) {
int count = 0;
ListNode *current = head;
while (current != NULL) {
current = current->next;
count++;
}
return count;
}
// Function to divide the list into two segments.
ListNode *divideList(ListNode *head, int size, ListNode **remainder) {
ListNode *prevMid = head;
ListNode *end = head->next;
// Find middle using fast and slow pointers approach
for (int i = 1; i < size && (prevMid->next != NULL || (end != NULL && end->next != NULL)); i++) {
if (end != NULL && end->next != NULL) {
end = (end->next->next != NULL) ? end->next->next : end->next;
}
if (prevMid->next != NULL) {
prevMid = prevMid->next;
}
}
ListNode *middle = prevMid->next;
prevMid->next = NULL;
*remainder = end != NULL ? end->next : NULL;
if (end != NULL) end->next = NULL;
return middle;
}
// Function for merging two lists into one in sorted order.
void mergeLists(ListNode **tailPointer, ListNode *first, ListNode *second) {
ListNode dummy;
ListNode *tail = &dummy;
dummy.next = NULL;
while (first != NULL && second != NULL) {
if (first->val < second->val) {
tail->next = first;
first = first->next;
} else {
tail->next = second;
second = second->next;
}
tail = tail->next;
}
tail->next = (first != NULL) ? first : second;
while (tail->next != NULL) {
tail = tail->next;
}
(*tailPointer)->next = dummy.next;
*tailPointer = tail;
}
// Function to sort the linked list using merge sort.
struct ListNode *mergeSort(struct ListNode *head) {
if (head == NULL || head->next == NULL) return head;
int total = getNodeCount(head);
ListNode *start = head;
ListNode dummy;
dummy.next = NULL;
ListNode *tail, *nextPart;
for (int size = 1; size < total; size *= 2) {
tail = &dummy;
while (start != NULL) {
if (start->next == NULL) {
tail->next = start;
break;
}
ListNode *midPoint = divideList(start, size, &nextPart);
mergeLists(&tail, start, midPoint);
start = nextPart;
}
start = dummy.next;
}
return dummy.next;
}
Sort a linked list efficiently using merge sort implemented in C. This code defines the operation of merging in the context of singly linked lists.
Key functions utilized:
- getNodeCount: Counts the nodes in the list.
- divideList: Splits the list into two parts based on the given size and updates the remainder for the next segment.
- mergeLists: Combines two lists into a single sorted list.
- mergeSort: Applies the merge sort algorithm on the linked list.
Steps include:
- First, determine the count of nodes in the linked list using
getNodeCount. - Split the list into two halves at each level of recursion using
divideList. - Merge these halves in a sorted manner using
mergeLists. - This process continues recursively with doubled segment size at each step until the entire list is sorted.
The algorithm uses a dummy node for simplicity in handling pointer references during the merging phase. The merge process adjusts next pointers to form a new sorted linked list step by step, avoiding complex data manipulations. The approach ensures stability and efficient sorting with time complexity typically O(n log n), where n is the number of nodes in the list. This mergesort technique effectively manages large lists compared to traditional sorting methods which can be slow and inefficient for linked structures.
var mergeSortList = function(listHead) {
if (!listHead || !listHead.next) return listHead;
function listLength(head) {
let count = 0,
current = head;
while (current) {
current = current.next;
count++;
}
return count;
}
function divideList(start, cutSize) {
let preMiddle = start;
let endPtr = start.next;
for (
let step = 1;
step < cutSize && (preMiddle.next || endPtr.next);
step++
) {
if (endPtr.next) {
endPtr = endPtr.next.next ? endPtr.next.next : endPtr.next;
}
if (preMiddle.next) {
preMiddle = preMiddle.next;
}
}
const middle = preMiddle.next;
preMiddle.next = null;
nextPart = endPtr.next;
endPtr.next = null;
return middle;
}
function mergeTwoLists(first, second) {
let tempHead = new ListNode();
let endOfList = tempHead;
while (first && second) {
if (first.val < second.val) {
endOfList.next = first;
first = first.next;
} else {
endOfList.next = second;
second = second.next;
}
endOfList = endOfList.next;
}
endOfList.next = first || second;
while (endOfList.next) {
endOfList = endOfList.next;
}
mergeTail.next = tempHead.next;
mergeTail = endOfList;
}
let totalNodes = listLength(listHead);
let subListStart = listHead;
let tempHead = new ListNode();
let mergeTail, nextPart;
for (let segSize = 1; segSize < totalNodes; segSize *= 2) {
mergeTail = tempHead;
while (subListStart) {
if (!subListStart.next) {
mergeTail.next = subListStart;
break;
}
let middle = divideList(subListStart, segSize);
mergeTwoLists(subListStart, middle);
subListStart = nextPart;
}
subListStart = tempHead.next;
}
return tempHead.next;
};
The provided JavaScript function mergeSortList implements the merge sort algorithm to sort a singly linked list. Follow these instructions to understand and utilize the code.
- Ensure the input to
mergeSortListis the head of a singly linked list. If the list is empty or has a single node, the function directly returns the head as no sorting is needed. - Understand that the core utility functions include:
listLength: calculates the total number of nodes in the list.divideList: divides the list into two halves based on a specified size, and manages the linking pointers correctly.mergeTwoLists: merges two sorted halves into a single sorted list.
- The
mergeSortListfunction uses an iterative approach:- Starts by determining the total node count using
listLength. - It then creates a temporary head node to simplify edge cases while merging lists.
- The sorting is done in multiple passes. For each pass, it doubles the segment size (starting from 1), which dictates how lists are divided and merged.
- Starts by determining the total node count using
- During each pass:
- The list is divided based on the segment size using
divideList. - The two divided lists are then merged back using
mergeTwoLists. - The process continues until the entire list is sorted in a single segment.
- The list is divided based on the segment size using
- Remember to maintain proper pointers to manage the next parts of the list and merged segments to ensure all parts of the list are sorted and recombined correctly.
By following this method, your linked list gets sorted efficiently using the divide-and-conquer principle of merge sort, making it suitable for large datasets where efficiency is crucial. The space complexity is constant, and the time complexity is O(n log n), which is optimal for sorting linked list data structures.
class Solution:
def sortLinkedList(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
total_len = self.getListCount(head)
current_head = head
dummy_head_node = ListNode()
current_size = 1
while current_size < total_len:
self.current_tail = dummy_head_node
while current_head:
if not current_head.next:
self.current_tail.next = current_head
break
middle = self.divide(current_head, current_size)
self.combine(current_head, middle)
current_head = self.next_part
current_head = dummy_head_node.next
current_size *= 2
return dummy_head_node.next
def divide(self, start, size):
mid_prev = start
tail = start.next
for _ in range(1, size):
if tail and tail.next:
tail = tail.next.next
elif tail:
tail = tail.next
if mid_prev.next:
mid_prev = mid_prev.next
mid_point = mid_prev.next
mid_prev.next = None
self.next_part = tail.next if tail else None
if tail:
tail.next = None
return mid_point
def combine(self, list1, list2):
dummy = ListNode()
tail = dummy
while list1 and list2:
if list1.val < list2.val:
tail.next = list1
list1 = list1.next
else:
tail.next = list2
list2 = list2.next
tail = tail.next
tail.next = list1 if list1 else list2
while tail.next:
tail = tail.next
self.current_tail.next = dummy.next
self.current_tail = tail
def getListCount(self, head):
length = 0
while head:
head = head.next
length += 1
return length
The provided Python code illustrates a method to sort a linked list using a bottom-up merge sort approach. Here's a breakdown of the implementation:
- The
sortLinkedListfunction serves as the main function that oversees the sorting process. It initializes by checking if the list is empty or contains only one element (base cases), in which case no sorting is needed. - A
dummy_head_nodeis used to facilitate easier handling of list nodes during merging phases without worrying about null-pointer exceptions. - The sort operates by recursively breaking down the list into smaller parts using the
dividefunction until the pieces can no longer be divided (basic divide-and-conquer algorithm). - The
combinefunction merges these parts in a sorted manner. This function uses adummynode to handle edge cases seamlessly and combines two halves while maintaining order. - A notable aspect of the function's efficiency is "bottom-up" approach, where the list size doubles each iteration reducing the logarithmic time complexity.
- The
getListCountfunction calculates the total number of elements in the list, necessary for the algorithm's operation.
Utilize this method for efficient and effective sorting of linked lists, especially when looking to maintain an order without additional space complexity introduced by methods like quicksort or heapsort in linked list structures. This implementation assures a time complexity of O(n log n) where n is the number of elements in the list.
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