0143. Reorder List (M)

题目

Given a singly linked list L: ,
reorder it to:

You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

题意

将给定链表按照指定顺序重新排列为新链表。

思路

利用快慢指针找到中间结点，将链表分为左右两部分，将右链表逆序后逐个插入左链表的间隔中。

代码实现

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void reorderList(ListNode head) {
        if (head == null) {
            return;
        }
        // 找到中间结点
        ListNode slow = head, fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode rHead = null;
        ListNode p = slow.next;
        slow.next = null;                // 注意断开左右链表，不然可能生成环
           // 逆置右链表
        while (p != null) {
            ListNode temp = p.next;
            p.next = rHead;
            rHead = p;
            p = temp;
        }
        // 将逆置后的右链表逐个插入左链表
        while (rHead != null) {
            ListNode temp = rHead.next;
            rHead.next = head.next;
            head.next = rHead;
            rHead = temp;
            head = head.next.next;
        }
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {void} Do not return anything, modify head in-place instead.
 */
var reorderList = function (head) {
  if (!head) return
  let slow = head
  let fast = head
  while (fast.next && fast.next.next) {
    fast = fast.next.next
    slow = slow.next
  }
  let head2 = null
  let cur = slow.next
  slow.next = null
  while (cur) {
    let tmp = cur.next
    cur.next = head2
    head2 = cur
    cur = tmp
  }
  while (head2) {
    let tmp = head2.next
    head2.next = head.next
    head.next = head2
    head = head2.next
    head2 = tmp
  }
}