Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes' values.
For example,
Given {1,2,3,4}
, reorder it to {1,4,2,3}
.
题目思路比较直接:
1)把整个链表划分成2个等长的子链表,如果原链表长度为奇数,那么第一个子链表的长度多1。
2)翻转第二个子链表;
3)将两个子链表合并。
代码里有些变量可以省去,为了看起来逻辑清晰,还是保留了。整个链表遍历了3次,但是没有使用额外空间。
public static void reorderList(ListNode head) { if (head == null || head.next == null) return; // partition the list into 2 sublists of equal length ListNode slowNode = head, fastNode = head; while (fastNode.next != null) { fastNode = fastNode.next; if (fastNode.next != null) { fastNode = fastNode.next; } else { break; } slowNode = slowNode.next; } // 2 sublist heads ListNode head1 = head, head2 = slowNode.next; // detach the two sublists slowNode.next = null; // reverse the second sublist ListNode cur = head2, post = cur.next; cur.next = null; while (post != null) { ListNode temp = post.next; post.next = cur; cur = post; post = temp; } head2 = cur; // the new head of the reversed sublist // merge the 2 sublists as required ListNode p = head1, q = head2; while (q != null) { ListNode temp1 = p.next; ListNode temp2 = q.next; p.next = q; q.next = temp1; p = temp1; q = temp2; } }