92. 反转链表 II

题目描述

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 <= 500
• 1 <= left <= right <= n

进阶： 你可以使用一趟扫描完成反转吗？

方法一：模拟

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 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为链表的长度。

Java

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;
    }
}

C++

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;
    }
};

TypeScript

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;
}

Python

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