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.
For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
if (head == null) {
return null;
}
ListNode dummy1 = new ListNode(0);
dummy1.next = head;
ListNode curt1 = dummy1;
ListNode dummy2 = new ListNode(0);
ListNode curt2 = dummy2;
while (curt1.next != null) {
if (curt1.next.val >= x) {
//update curt1
ListNode removedNode = curt1.next;
curt1.next = curt1.next.next;
//update curt2
curt2.next = removedNode;
curt2 = curt2.next;
// make it detached from original list
curt2.next = null;
} else {
curt1 = curt1.next;
}
}
curt1.next = dummy2.next;
return dummy1.next;
}
}
