首页 > 代码库 > Leetcode: Reorder List
Leetcode: Reorder List
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}.
这是一道比较综合的链表操作的题目,要按照题目要求给链表重新连接成要求的结果。其实理清思路也比较简单,分三步完成:(1)将链表切成两半,也就是找到中点,然后截成两条链表;(2)将后面一条链表进行reverse操作,就是反转过来;(3)将两条链表按顺序依次merge起来。
这几个操作都是我们曾经接触过的操作了,第一步找中点就是用runner technique方法,一个两倍速跑,一个一倍速跑,知道快的碰到链表尾部,慢的就正好停在中点了。第二步是比较常见的reverse操作,在Reverse Nodes in k-Group也有用到了,一般就是一个个的翻转过来即可。第三步是一个merge操作,做法类似于Sort List中的merge
接下来看看时间复杂度,第一步扫描链表一遍,是O(n),第二步对半条链表做一次反转,也是O(n),第三部对两条半链表进行合并,也是一遍O(n)。所以总的时间复杂度还是O(n),由于过程中没有用到额外空间,所以空间复杂度O(1)。代码如下:
1 /** 2 * Definition for singly-linked list. 3 * class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public class Solution {13 public void reorderList(ListNode head) {14 if (head == null || head.next == null) return;15 ListNode dummy = new ListNode(-1);16 dummy.next = head;17 ListNode current = dummy;18 ListNode runner = dummy;19 while (runner.next != null && runner.next.next != null) {20 current = current.next;21 runner = runner.next.next;22 }23 ListNode half = current.next;24 current.next = null;25 26 /*reverse the second Linked List*/27 if (runner.next != null) runner = runner.next; //make sure the runner pointer points to the end of the the second Linked List28 ListNode dummy2 = new ListNode(-2);29 dummy2.next = half;//create another dummy node whose next points to the head of the second Linked List30 while (dummy2.next != runner) {31 ListNode dnext = dummy2.next.next;32 ListNode rnext = runner.next;33 dummy2.next.next = rnext;34 runner.next = dummy2.next;35 dummy2.next = dnext;36 }37 38 /*merge the two Linked List together*/39 ListNode header1 = dummy;40 ListNode header2 = dummy2;41 while (header1.next != null && header2.next != null) {42 ListNode store = header1.next.next;43 ListNode merge = new ListNode(header2.next.val);44 header1.next.next = merge;45 merge.next = store;46 header1 = header1.next.next;47 header2 = header2.next;48 }49 if (header2.next != null) {50 header1.next = header2.next;51 }52 }53 }
Leetcode: Reorder List
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。