leetcode 81: Partition List

Partition ListApr
30 '12

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;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
// Start typing your Java solution below
// DO NOT write main() function
if( head==null) return head;

ListNode less = new ListNode(-1);
ListNode greater = new ListNode(-1);

ListNode p=less, q=greater;

while(head!=null) {
if(head.val<x) {
p.next = head;
head = head.next;
p = p.next;
p.next = null;
} else {
q.next = head;
head = head.next;
q = q.next;
q.next = null;
}

}

p.next = greater.next;
return less.next;
}
}