README.md

92. 反转链表 II

English Version

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

Python3

# 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: ListNode, left: int, right: int) -> ListNode:
        if head is None or 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

Java

/**
 * 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 == null || 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;
    }
}

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
 * @param {number} left
 * @param {number} right
 * @return {ListNode}
 */
var reverseBetween = function (head, left, right) {
    if (!head || !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;
};

C++

/**
 * 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 == nullptr || head->next == nullptr || 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;
    }
};

C#

/**
 * 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 == null || 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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