This content originally appeared on DEV Community and was authored by seanpgallivan
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Leetcode Problem #86 (Medium): Partition List
Description:
(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)
Given the
headof a linked list and a valuex, partition it such that all nodes less thanxcome before nodes greater than or equal tox.You should preserve the original relative order of the nodes in each of the two partitions.
Examples:
Example 1: Input: head = [1,4,3,2,5,2], x = 3 Output: [1,2,2,4,3,5] Visual:
Example 2: Input: head = [2,1], x = 2 Output: [1,2]
Constraints:
- The number of nodes in the list is in the range
[0, 200].-100 <= Node.val <= 100-200 <= x <= 200
Idea:
(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)
The easiest thing to do here is to create separate linked lists for the front and back portions of list we want to return. In order to accomplish that, we should first create some dummy heads (fdum, bdum), then create pointers for the current nodes each of the front, back, and main lists (front, back, curr).
Then we can simply iterate through the main list and stitch together each node to either front or back, depending on the node's value.
Once we reach the end, we just need to stitch together the two sub-lists, making sure to cap off the end of back, and then return our new list, minus the dummy head.
Implementation:
There are only minor differences between the code of all four languages.
Javascript Code:
(Jump to: Problem Description || Solution Idea)
var partition = function(head, x) {
let fdum = new ListNode(0), bdum = new ListNode(0),
front = fdum, back = bdum, curr = head
while (curr) {
if (curr.val < x)front.next = curr, front = curr
else back.next = curr, back = curr
curr = curr.next
}
front.next = bdum.next, back.next = null
return fdum.next
};
Python Code:
(Jump to: Problem Description || Solution Idea)
class Solution:
def partition(self, head: ListNode, x: int) -> ListNode:
fdum, bdum = ListNode(0), ListNode(0)
front, back, curr = fdum, bdum, head
while curr:
if curr.val < x:
front.next = curr
front = curr
else:
back.next = curr
back = curr
curr = curr.next
front.next, back.next = bdum.next, None
return fdum.next
Java Code:
(Jump to: Problem Description || Solution Idea)
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode fdum = new ListNode(0), bdum = new ListNode(0),
front = fdum, back = bdum, curr = head;
while (curr != null) {
if (curr.val < x) {
front.next = curr;
front = curr;
} else {
back.next = curr;
back = curr;
}
curr = curr.next;
}
front.next = bdum.next;
back.next = null;
return fdum.next;
}
}
C++ Code:
(Jump to: Problem Description || Solution Idea)
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode *fdum = new ListNode(0), *bdum = new ListNode(0),
*front = fdum, *back = bdum, *curr = head;
while (curr) {
if (curr->val < x) front->next = curr, front = curr;
else back->next = curr, back = curr;
curr = curr->next;
}
front->next = bdum->next, back->next = nullptr;
return fdum->next;
}
};
This content originally appeared on DEV Community and was authored by seanpgallivan
seanpgallivan | Sciencx (2021-04-14T07:58:52+00:00) Solution: Partition List. Retrieved from https://www.scien.cx/2021/04/14/solution-partition-list/
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