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题目(Reorder List)
链接：https://leetcode-cn.com/problems/reorder-list
解决数：1386
通过率：63.6%
标签：栈 递归 链表 双指针
相关公司：bytedance facebook amazon
给定一个单链表 L **的头节点 head ，单链表 L 表示为：
L0 → L1 → … → Ln - 1 → Ln
请将其重新排列后变为：
L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …
不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
示例 1：
输入： head = [1,2,3,4]
输出： [1,4,2,3]
示例 2：
输入： head = [1,2,3,4,5]
输出： [1,5,2,4,3]
提示：
- 链表的长度范围为
[1, 5 * 104] 1 <= node.val <= 1000
思路
- 步骤一: 找到链表中点后分割其为 left 链表、right 链表两部分;
- 步骤二: 翻转 right 链表, 翻转链表思路同 206.Reverse_Linked_List;
- 步骤三: 接着从 left 链表的左侧, 翻转后的 right 链表的左侧各取一个值进行交替拼接;
快慢指针即 quick 指针每次走两步, slow 指针每次走一步, 同 148.Sort_List
s q
dummy -> 1 -> 2 -> 3 -> 4 -> 5 -> NULL
步骤一: 拆分 head 链表成 `left`、`right` 两个链表;
l
1 -> 2 -> 3 -> null
r
4 -> 5 -> NULL
步骤二: 翻转 right 链表
l
1 -> 2 -> 3 -> null
r
5 -> 4 -> NULL
步骤三: 衔接 left、right 两个链表;
l
1 -> 5 -> 2 -> 3 -> NULL
r
4 -> NULL
步骤二中翻转链表的图解大致如下:
cur next
1 -> 2 -> 3 -> null
prev cur
null <- 1 <- 2 -> 3 -> null
prev cur
null <- 1 <- 2 <- 3 -> null
prev
null <- 1 <- 2 <- 3 -> null
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {void} Do not return anything, modify head in-place instead.
*/
var reorderList = function(head) {
const dummy = new ListNode(0)
dummy.next = head
let slow = dummy
let quick = dummy
while (quick && quick.next) {
slow = slow.next
quick = quick.next
quick = quick.next
}
let right = slow.next
slow.next = null
let left = dummy.next
right = reverseList(right)
while (left && right) {
let lNext = left.next
let rNext = right.next
right.next = left.next
left.next = right
left = lNext
right = rNext
}
return dummy.next
}
var reverseList = (list) => {
let prev = null
let cur = list
while (cur) {
let next = cur.next
cur.next = prev
prev = cur
cur = next
}
return prev
}
