Given the head of a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
Example 1:
Input: head = [1,4,3,2,5,2], x = 3 Output: [1,2,2,4,3,5]
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
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def partition(self, head: ListNode, x: int) -> ListNode:
l1, l2 = ListNode(), ListNode()
cur1, cur2 = l1, l2
while head:
if head.val < x:
cur1.next = head
cur1 = cur1.next
else:
cur2.next = head
cur2 = cur2.next
head = head.next
cur1.next = l2.next
cur2.next = None
return l1.next/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode l1 = new ListNode(0);
ListNode l2 = new ListNode(0);
ListNode cur1 = l1, cur2 = l2;
while (head != null) {
if (head.val < x) {
cur1.next = head;
cur1 = cur1.next;
} else {
cur2.next = head;
cur2 = cur2.next;
}
head = head.next;
}
cur1.next = l2.next;
cur2.next = null;
return l1.next;
}
}/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode* l1 = new ListNode();
ListNode* l2 = new ListNode();
ListNode* cur1 = l1;
ListNode* cur2 = l2;
while (head != nullptr) {
if (head->val < x) {
cur1->next = head;
cur1 = cur1->next;
} else {
cur2->next = head;
cur2 = cur2->next;
}
head = head->next;
}
cur1->next = l2->next;
cur2->next = nullptr;
return l1->next;
}
};/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} x
* @return {ListNode}
*/
var partition = function (head, x) {
if (!head || !head.next) {
return head;
}
const dummy1 = new ListNode(-1), dummy2 = new ListNode(-1);
let cur1 = dummy1, cur2 = dummy2;
let cur = head;
while (cur) {
if (cur.val < x) {
cur1.next = cur;
cur1 = cur1.next;
} else {
cur2.next = cur;
cur2 = cur2.next;
}
cur = cur.next;
}
cur2.next = null;
cur1.next = dummy2.next;
return (dummy1.next);
};