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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 int val;
3 ListNode next;
4 ListNode(int x) { val = x; }
5}
6
7public ListNode[] splitListToParts(ListNode root, int k) {
8 ListNode[] parts = new ListNode[k];
9 int length = 0;
10 ListNode node = root;
11 while (node != null) {
12 length++;
13 node = node.next;
14 }
15 int partSize = length / k;
16 int extraNodes = length % k;
17 node = root;
18 for (int i = 0; i < k && node != null; i++) {
19 parts[i] = node;
20 int currentPartSize = partSize + (i < extraNodes ? 1 : 0);
21 for (int j = 1; j < currentPartSize; j++) {
22 node = node.next;
23 }
24 ListNode nextPart = node.next;
25 node.next = null;
26 node = nextPart;
27 }
28 return parts;
29}In this Java solution, the linked list's length is calculated first to determine how to split it into k parts. An array of ListNode is used to store the resulting parts. The algorithm ensures that earlier parts can be larger by calculating the appropriate sizes using integer division and remainder.
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.