LeetCode Solutions
86. Partition List
Time: $O(n)$ Space: $O(1)$
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode beforeHead(0);
ListNode afterHead(0);
ListNode* before = &beforeHead;
ListNode* after = &afterHead;
for (; head; head = head->next)
if (head->val < x) {
before->next = head;
before = head;
} else {
after->next = head;
after = head;
}
after->next = nullptr;
before->next = afterHead.next;
return beforeHead.next;
};
};
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode beforeHead = new ListNode(0);
ListNode afterHead = new ListNode(0);
ListNode before = beforeHead;
ListNode after = afterHead;
for (; head != null; head = head.next)
if (head.val < x) {
before.next = head;
before = head;
} else {
after.next = head;
after = head;
}
after.next = null;
before.next = afterHead.next;
return beforeHead.next;
}
}
class Solution:
def partition(self, head: ListNode, x: int) -> ListNode:
beforeHead = ListNode(0)
afterHead = ListNode(0)
before = beforeHead
after = afterHead
while head:
if head.val < x:
before.next = head
before = head
else:
after.next = head
after = head
head = head.next
after.next = None
before.next = afterHead.next
return beforeHead.next