Leetcode 86. 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.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

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Analysis:

1. build two empty lists
2. val < x add it to the tail of first list
    val >= x add it to the tail of second list
3. concatenate the tail of list1 to the head of list2
4. the tail of list2 list.next = null

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Java code:

20160601

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        //1. build two empty lists
        //2. val < x add it to the tail of first list
        //   val >= x add it to the tail of second list
        //3. concatenate the tail of list1 to the head of list2
        //4. the tail of list2 list.next = null
        
        //base case
        if(head == null || head.next == null) {
            return head;
        }
        
        ListNode cur = head;
        ListNode dummy1 = new ListNode(0);
        ListNode head1 = dummy1;
        ListNode dummy2 = new ListNode(0);
        ListNode head2 = dummy2;
        while( cur != null) {
            if (cur.val < x) {
                dummy1.next = cur;
                dummy1 = dummy1.next;
            } else {
                dummy2.next = cur;  
                dummy2 = dummy2.next;
            }
            cur = cur.next;
        }
        dummy2.next = null;
        dummy1.next = head2.next;
        return head1.next;
    }
}