Problem
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
Solution
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode dummy1 = new ListNode(0);
ListNode dummy2 = new ListNode(0);
ListNode small = dummy1, large = dummy2;
while (head != null) {
if (head.val < x) {
small.next = head;
small = small.next;
} else {
large.next = head;
large = large.next;
}
head = head.next;
}
small.next = dummy2.next;
large.next = null;
return dummy1.next;
}
}
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