| comments | difficulty | edit_url | tags | |
|---|---|---|---|---|
true |
中等 |
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给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。
示例 1：
输入：head = [1,2,3,4,5], left = 2, right = 4 输出：[1,4,3,2,5]
示例 2：
输入：head = [5], left = 1, right = 1 输出：[5]
提示：
- 链表中节点数目为
n 1 <= n <= 500-500 <= Node.val <= 5001 <= left <= right <= n
进阶： 你可以使用一趟扫描完成反转吗？
定义一个虚拟头结点 dummy，指向链表的头结点 head，然后定义一个指针 pre 指向 dummy，从虚拟头结点开始遍历链表，遍历到第 left 个结点时，将 pre 指向该结点，然后从该结点开始遍历 right - left + 1 次，将遍历到的结点依次插入到 pre 的后面，最后返回 dummy.next 即可。
时间复杂度
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseBetween(
self, head: Optional[ListNode], left: int, right: int
) -> Optional[ListNode]:
if head.next is None or left == right:
return head
dummy = ListNode(0, head)
pre = dummy
for _ in range(left - 1):
pre = pre.next
p, q = pre, pre.next
cur = q
for _ in range(right - left + 1):
t = cur.next
cur.next = pre
pre, cur = cur, t
p.next = pre
q.next = cur
return dummy.next/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseBetween(ListNode head, int left, int right) {
if (head.next == null || left == right) {
return head;
}
ListNode dummy = new ListNode(0, head);
ListNode pre = dummy;
for (int i = 0; i < left - 1; ++i) {
pre = pre.next;
}
ListNode p = pre;
ListNode q = pre.next;
ListNode cur = q;
for (int i = 0; i < right - left + 1; ++i) {
ListNode t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
p.next = pre;
q.next = cur;
return dummy.next;
}
}/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int left, int right) {
if (!head->next || left == right) {
return head;
}
ListNode* dummy = new ListNode(0, head);
ListNode* pre = dummy;
for (int i = 0; i < left - 1; ++i) {
pre = pre->next;
}
ListNode *p = pre, *q = pre->next;
ListNode* cur = q;
for (int i = 0; i < right - left + 1; ++i) {
ListNode* t = cur->next;
cur->next = pre;
pre = cur;
cur = t;
}
p->next = pre;
q->next = cur;
return dummy->next;
}
};/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseBetween(head *ListNode, left int, right int) *ListNode {
if head.Next == nil || left == right {
return head
}
dummy := &ListNode{0, head}
pre := dummy
for i := 0; i < left-1; i++ {
pre = pre.Next
}
p, q := pre, pre.Next
cur := q
for i := 0; i < right-left+1; i++ {
t := cur.Next
cur.Next = pre
pre = cur
cur = t
}
p.Next = pre
q.Next = cur
return dummy.Next
}/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function reverseBetween(head: ListNode | null, left: number, right: number): ListNode | null {
const n = right - left;
if (n === 0) {
return head;
}
const dummy = new ListNode(0, head);
let pre = null;
let cur = dummy;
for (let i = 0; i < left; i++) {
pre = cur;
cur = cur.next;
}
const h = pre;
pre = null;
for (let i = 0; i <= n; i++) {
const next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
h.next.next = cur;
h.next = pre;
return dummy.next;
}// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
// pub val: i32,
// pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
// #[inline]
// fn new(val: i32) -> Self {
// ListNode {
// next: None,
// val
// }
// }
// }
impl Solution {
pub fn reverse_between(
head: Option<Box<ListNode>>,
left: i32,
right: i32,
) -> Option<Box<ListNode>> {
let mut dummy = Some(Box::new(ListNode { val: 0, next: head }));
let mut pre = &mut dummy;
for _ in 1..left {
pre = &mut pre.as_mut().unwrap().next;
}
let mut cur = pre.as_mut().unwrap().next.take();
for _ in 0..right - left + 1 {
let mut next = cur.as_mut().unwrap().next.take();
cur.as_mut().unwrap().next = pre.as_mut().unwrap().next.take();
pre.as_mut().unwrap().next = cur.take();
cur = next;
}
for _ in 0..right - left + 1 {
pre = &mut pre.as_mut().unwrap().next;
}
pre.as_mut().unwrap().next = cur;
dummy.unwrap().next
}
}/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} left
* @param {number} right
* @return {ListNode}
*/
var reverseBetween = function (head, left, right) {
if (!head.next || left == right) {
return head;
}
const dummy = new ListNode(0, head);
let pre = dummy;
for (let i = 0; i < left - 1; ++i) {
pre = pre.next;
}
const p = pre;
const q = pre.next;
let cur = q;
for (let i = 0; i < right - left + 1; ++i) {
const t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
p.next = pre;
q.next = cur;
return dummy.next;
};/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution {
public ListNode ReverseBetween(ListNode head, int left, int right) {
if (head.next == null || left == right) {
return head;
}
ListNode dummy = new ListNode(0, head);
ListNode pre = dummy;
for (int i = 0; i < left - 1; ++i) {
pre = pre.next;
}
ListNode p = pre;
ListNode q = pre.next;
ListNode cur = q;
for (int i = 0; i < right - left + 1; ++i) {
ListNode t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
p.next = pre;
q.next = cur;
return dummy.next;
}
}