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？

解法

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。

Python3JavaC++GoTypeScriptRustJavaScriptC#

# 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;
    }
}

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