86. Partition List
Given 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:
Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5
public ListNode partition(ListNode head, int x) {
ListNode sentinel = new ListNode(0);
sentinel.next = head;
ListNode boundary = sentinel, prev = sentinel, curr = head;
while(curr != null) {
if(curr.val < x && boundary != prev) {
prev.next = curr.next;
curr.next = boundary.next;
boundary.next = curr;
boundary = boundary.next;
curr = prev.next;
} else {
if(curr.val < x && boundary == prev)
boundary = boundary.next;
prev = prev.next;
curr = curr.next;
}
}
return sentinel.next;
}