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.

这道题是一道分割的题，要求把比x小的元素放到x的前面，而且相对顺序不能变

这个时候我们采用双指针法

runner1用于寻找最后一个比x小得元素，这里作为插入点（分割点）

runner2用于寻找应该插到插入点之后的元素

所以这里会出现三种情况

情况1: runner2的元素小于x，这个时候如果runner1和runner2指向同一个元素，说明还没有找到分割点，所以两个指针继续往前走就行了。

情况2: runner2的元素小于x，这个时候如果runner1和runner2指向不同的元素，说明runner1已经走到了分割点前，而runner2.next就是应该被插到分割点前。把runner2.next插入到分割点前就ok

情况3: runner2的元素大于x, 这个时候找到了分割点，只移动runner2就可以了。直到runner2.next小于分割点。然后按照情况2来操作就可以了

情况1和情况2：

			if (runner2.next.val < x) {
				if (runner1 == runner2) {
					runner1 = runner1.next;
					runner2 = runner2.next;
				} else {
					ListNode temp = runner2.next;
					ListNode next = temp.next;
					temp.next = runner1.next;
					runner1.next = temp;
					runner2.next = next;
					runner1 = runner1.next;
				}
			} 

			else
				runner2 = runner2.next;

	public ListNode partition(ListNode head, int x) {
		if (head == null)
			return head;

		ListNode helper = new ListNode(0);
		helper.next = head;
		ListNode runner1 = helper;
		ListNode runner2 = helper;

		while (runner2 != null && runner2.next != null) {
			if (runner2.next.val < x) {
				if (runner1 == runner2) {
					runner1 = runner1.next;
					runner2 = runner2.next;
				} else {
					ListNode temp = runner2.next;
					ListNode next = temp.next;
					temp.next = runner1.next;
					runner1.next = temp;
					runner2.next = next;
					runner1 = runner1.next;
				}
			} else
				runner2 = runner2.next;
		}
		return helper.next;
	}

【Leetcode】Partition List (Swap)