README.md

92. 反转链表 II

English Version

题目描述

反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

说明:
1 ≤ m ≤ n ≤ 链表长度。

示例:

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

解法

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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