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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.
1function ListNode(val, next = null) {
2 this.val = val;
3 this.next = next;
4}
5
6var splitListToParts = function(root, k) {
7 const parts = new Array(k).fill(null);
8 let length = 0;
9 let node = root;
10 while (node) {
11 length++;
12 node = node.next;
13 }
14 const partSize = Math.floor(length / k);
15 let extraNodes = length % k;
16 node = root;
17 for (let i = 0; i < k && node; i++) {
18 parts[i] = node;
19 let currentSize = partSize + (i < extraNodes ? 1 : 0);
20 for (let j = 1; j < currentSize; j++) {
21 node = node.next;
22 }
23 let nextPart = node.next;
24 node.next = null;
25 node = nextPart;
26 }
27 return parts;
28};In JavaScript, the solution maintains a similar structure: calculating the linked list length and dividing it into k parts. An array is used to store the head of each list part. Integer operations and logic handle the distribution of nodes.
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.