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; } }