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->3->4->5.

题意：有一个无序的链表，给了一个值x，要把链表中小于x的值排列在大于等于x的值之前，并且要维持原链表的顺序。

思路：题意相当于把链表分成两部分，小于x的和大于等于x的两端。于是想到可以用两个新链表来记录这两段，最后把小于x那段的尾部和另一段的头部连结起来就可以了。

public ListNode partition(ListNode head, int x) {
    if (head == null || head.next == null) {
        return head;
    }

    ListNode left = new ListNode(0);
    ListNode leftRoot = left;
    ListNode right = new ListNode(0);
    ListNode rightRoot = right;
    ListNode dummy = head;
    while (dummy != null) {
        if (dummy.val < x) {
            left.next = dummy;
            left = left.next;
        } else {
            right.next = dummy;
            right = right.next;
        }
        dummy = dummy.next;
    }
    left.next = null;
    right.next = null;

    left.next = rightRoot.next;

    return leftRoot.next;
}