【Leetcode】Partition List (Swap)

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.

这道题是一道切割的题。要求把比x小的元素放到x的前面。并且相对顺序不能变

这个时候我们採用双指针法

runner1用于寻找最后一个比x小得元素，这里作为插入点（切割点）

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。这个时候假设runner1和runner2指向同一个元素。说明还没有找到切割点，所以两个指针继续往前走即可了。

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

情况3: runner2的元素大于x, 这个时候找到了切割点，仅仅移动runner2就能够了。直到runner2.next小于切割点。

然后依照情况2来操作就能够了

情况1和情况2：

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

情况3：

			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 insert = runner2.next;
					ListNode next = insert.next;
					insert.next = runner1.next;
					runner1.next = insert;
					runner2.next = next;
					runner1 = runner1.next;
				}
			} else
				runner2 = runner2.next;
		}
		return helper.next;
	}