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This approach involves two main steps. First, traverse the linked list to determine its length. Then, decide how to split this length evenly across the k parts. Calculate the base size for each part and determine how many parts need an extra node.
Time Complexity: O(n) where n is the length of the linked list.
Space Complexity: O(k) for storing the resulting list parts.
1class ListNode:
2 def __init__(self, x):
3 self.val = x
4 self.next = None
5
6class Solution:
7 def splitListToParts(self, root: ListNode, k: int):
8 parts = [None] * k
9 length = 0
10 node = root
11 while node:
12 length += 1
13 node = node.next
14 partSize = length // k
15 extraNodes = length % k
16 node = root
17 for i in range(k):
18 parts[i] = node
19 currentPartSize = partSize + (1 if i < extraNodes else 0)
20 for j in range(currentPartSize - 1):
21 if node:
22 node = node.next
23 if node:
24 nextPart = node.next
25 node.next = None
26 node = nextPart
27 return partsThis Python solution follows the same logic: it calculates the length of the linked list, decides on the split sizes for each part, and uses list indexing to store the result. Python’s handling of list operations and iterations makes the code concise.
This approach involves iterating through the linked list and modifying the pointers to split the list directly into parts of calculated sizes based on total length and k. It ensures that the list splitting does not require additional passes, combining calculation and splitting in a single traversal.
Time Complexity: O(n) where n is the length of the linked list.
Space Complexity: O(k) for storing the resulting list pointers.
1 public int val;
public ListNode next;
public ListNode(int val = 0, ListNode next = null) {
this.val = val;
this.next = next;
}
}
public class Solution {
public ListNode[] SplitListToParts(ListNode root, int k) {
ListNode[] parts = new ListNode[k];
int length = 0;
ListNode node = root;
while (node != null) {
length++;
node = node.next;
}
int partSize = length / k, extraNodes = length % k;
node = root;
for (int i = 0; i < k && node != null; i++) {
parts[i] = node;
int currentSize = partSize + (i < extraNodes);
for (int j = 1; j < currentSize; j++) {
node = node.next;
}
if (node != null) {
ListNode nextPart = node.next;
node.next = null;
node = nextPart;
}
}
return parts;
}
}This C# implementation efficiently traverses the list once, dividing it by correctly shifting the next pointers. It employs simple logic to determine where to make the cuts, avoiding further passes.