Partition List

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

巧妙的利用dummy node

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head == null){
            return head;
        }
        
        ListNode leftDummy = new ListNode(-1);
        ListNode rightDummy = new ListNode(-1);
        ListNode right = rightDummy, left = leftDummy;
        
        while(head != null){
            if(head.val < x){
                left.next = head;
                left = head;
            }else{
                right.next = head;
                right = head;
            }
            head = head.next;
        }
        
        
        right.next = null;
        left.next = rightDummy.next;
        
        return leftDummy.next;
    }
}