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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5
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Related Topics 
Linked ListTwo Pointers
Java	





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/**
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 * Definition for singly-linked list.
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 * public class ListNode {
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 *     int val;
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 *     ListNode next;
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 *     ListNode(int x) { val = x; }
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 * }
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 */
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class Solution {
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    public ListNode partition(ListNode head, int x) {
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        if(head ==null){
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            return null;
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        }
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       ListNode largeHeadNode = null;
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        ListNode largeTailNode = null;
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        ListNode lessHeadNode = null;
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        ListNode lessTailNode  = null;
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        ListNode temp = head;
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        while(temp!=null){
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            if(temp.val >=x ){
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                if(largeTailNode ==null){
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                    largeHeadNode = temp;
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                    largeTailNode = temp;
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                }else{
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                    largeTailNode.next = temp;
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                    largeTailNode  = temp;
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                }
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            }else{
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                if(lessTailNode == null){
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                    lessTailNode = temp;
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                    lessHeadNode = temp;
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                }else{
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                    lessTailNode.next = temp;
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                    lessTailNode = temp;
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                }
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            }
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            temp = temp.next;
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        }
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        if(largeTailNode !=null){
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            largeTailNode.next = null;
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        }
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        if(lessTailNode !=null){
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            lessTailNode.next = largeHeadNode;
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        }
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        if(lessHeadNode == null){
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            return largeHeadNode;
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        }
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        return lessHeadNode;
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        }
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}