Reverse Linked List II

最新推荐文章于 2025-05-21 15:08:20 发布

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本文介绍了一种算法，用于在给定位置m到n之间反转单链表的部分元素，并且要求在原地完成反转操作，只需遍历一次链表。例如，对于链表1->2->3->4->5->NULL，当m=2且n=4时，反转后的链表为1->4->3->2->5->NULL。

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode newHead = new ListNode(0);
        newHead.next = head;
        
        ListNode prevNode = getMthNode(newHead, m);
        ListNode current = prevNode.next;
        ListNode tail = prevNode.next;
        int count = 1;
        
        ListNode prev = null;
        ListNode next = null;
        while (count <= (n - m + 1)) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
            count++;
        }
        prevNode.next = prev;
        tail.next = next;        
        
        return newHead.next;
    }
    
    public ListNode getMthNode(ListNode newHead, int m) {
        for (int i = 1; i < m; i++) {
            newHead = newHead.next;
        }
        return newHead;
    }
}