Problem

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

思路

• 其实最重要的, 首先拿到这一题, 第一眼最直接快捷的方法是什么?
  • 把小于target的移到左边, 大于等于target的放到右边.
• 而头指针可能会变, 所有要有两个dummy node.
• 组成两个链表后, 连接左链表末尾和右链表头

    public ListNode partition(ListNode head, int x) {
        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode left = leftDummy;
        ListNode right = rightDummy;

        while (head != null) {
            if (head.val < x) {
                left.next = head;
                left = left.next;
            } else {
                right.next = head;
                right = right.next;
            }
            head = head.next;
        }
        // 这步不能忘
        right.next = null;
        left.next = rightDummy.next;
        return leftDummy.next;
    }