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.
Solution:
/**
* 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) {
ListNode dummy1 = new ListNode(0);
ListNode dummy2 = new ListNode(0);
ListNode p1 = dummy1;
ListNode p2 = dummy2;
while(head!=null){
if(head.val<x){
p1.next = head;
p1 = p1.next;
} else {
p2.next = head;
p2 = p2.next;
}
head = head.next;
}
p1.next = dummy2.next;
p2.next = null;
return dummy1.next;
}
}