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.
public class Solution { public ListNode partition(ListNode head, int x) { // Start typing your Java solution below // DO NOT write main() function ListNode less = new ListNode(0); ListNode greater = new ListNode(0); ListNode result = less; ListNode pivot = greater; while(head!=null){ if(head.val < x){ less.next = new ListNode(head.val); less = less.next; } else{ greater.next = new ListNode(head.val); greater = greater.next; } head = head.next; } less.next = pivot.next; return result.next; } }
mistakes: when make new listnode(0) give value to listnode.next