README_EN.md

92. Reverse Linked List II

中文文档

Description

Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.

 

Example 1:

Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]

Example 2:

Input: head = [5], left = 1, right = 1
Output: [5]

 

Constraints:

• The number of nodes in the list is n.
• 1 <= n <= 500
• -500 <= Node.val <= 500
• 1 <= left <= right <= n

 

Follow up: Could you do it in one pass?

Solutions

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        if head is None or head.next is None or m == n:
            return head
        dummy = ListNode(0)
        dummy.next = head
        pre, cur = dummy, head
        for _ in range(m - 1):
            pre = cur
            cur = cur.next
        p, q = pre, cur
        for _ in range(n - m + 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(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head == null || head.next == null || m == n) {
            return head;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy, cur = head;
        for (int i = 0; i < m - 1; ++i) {
            pre = cur;
            cur = cur.next;
        }
        ListNode p = pre, q = cur;
        for (int i = 0; i < n - m + 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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