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Write code to partition a linked list around a value x, such that all nodes less than x come before all nodes greater than or equal to x. If x is contained within the list, the values of x only need to be after the elements less than x (see below). The partition element x can appear anywhere in the "right partition"; it does not need to appear between the left and right partitions.
Example:
Input: head = 3->5->8->5->10->2->1, x = 5 Output: 3->1->2->10->5->5->8
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def partition(self, head: ListNode, x: int) -> ListNode:
l1, l2 = ListNode(0), ListNode(0)
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;
}
};
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