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.

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

題意：

給定一個鏈表和一個值，把小於等於和大於該值的部分分別放到鏈表的一前一後去。

思路：

先分為兩個鏈表

然後合並

代碼：

 1 class Solution {
 2     public ListNode partition(ListNode head, int x) {
 3         ListNode leftDummy = new ListNode(-1);
 4         ListNode rightDummy = new ListNode (-1);    
 5         ListNode left_cur = leftDummy;
 6         ListNode right_cur = rightDummy; 
 
 7         ListNode cur = head;
 8         
 9         while( cur != null){
10             if(cur.val < x){
11                 left_cur.next = cur;
12                 left_cur = cur;
13             }else{
14                 right_cur.next = cur;
15                 right_cur = cur;
16             }
 
17             cur = cur.next;    
18         }
19        left_cur.next = rightDummy.next;
20        right_cur.next = null;
21        return leftDummy.next;       
22     }

[leetcode]86. Partition List劃分鏈表