【Leetcode】Reorder List (Walker_Runner)

这道题让把一串链表1,2,3...n-1,n变成1,n-1, 2, n-2

基本思想是把整个链表从中间切割成两半，然后把后边一半逆置，然后merge就可以了

代码如下

	public void reorderList(ListNode head) {
		if (head == null)
			return;
		ListNode helper = new ListNode(0);
		helper.next = head;
		ListNode walker = helper;
		ListNode runner = helper;
		while (runner != null && runner.next != null) {
			walker = walker.next;
			runner = runner.next.next;
		}
		ListNode head2 = walker.next;
		walker.next = null;
		head2 = reverse(head2);
		merge(head, head2);
	}

	public ListNode reverse(ListNode head) {
		if (head == null)
			return head;
		ListNode reverseHead = null;
		ListNode runner = head;
		while (runner != null) {
			ListNode next = runner.next;
			runner.next = reverseHead;
			reverseHead = runner;
			runner = next;
		}
		return reverseHead;
	}

	public void merge(ListNode l1, ListNode l2) {
		while (l1 != null && l2 != null) {
			if (l1.next == null) {
				l1.next = l2;
				return;
			}
			ListNode next = l2.next;
			l2.next = l1.next;
			l1.next = l2;
			l1 = l2.next;
			l2 = next;
		}
	}

merge的部分一定要先判断l1.next==null 的情况，不然后面会出现空指针，而且要记着return