LeetCode92 : Reverse Linked List II

Description

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

Note: 1 ≤ m ≤ n ≤ length of list.

Example

 Input: 1->2->3->4->5->NULL, m = 2, n = 4
 Output: 1->4->3->2->5->NULL

解题思路

这个是反转链表的升级版,反转的是链表中的一部分,但是原理基本一样,刚开始自己忽略了可以利用第一个节点的位置加上长度就可以确定最后一个反转的位置。还有一点需要补充:为什么需要重新构造一个虚拟头结点,这个是因为如果你要反转整个链表时,直接返回 head 会丢失一部分节点。这一点值得注意!!!

代码

下面这段代码来自于 LeetCode 最佳答案。

/**
 * 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) return null;
    ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list
    dummy.next = head;
    ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing
    for(int i = 0; i<m-1; i++) pre = pre.next;

    ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed
    ListNode then = start.next; // a pointer to a node that will be reversed

    // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3
    // dummy-> 1 -> 2 -> 3 -> 4 -> 5

    for(int i=0; i<n-m; i++)
    {
        start.next = then.next;
        then.next = pre.next;
        pre.next = then;
        then = start.next;
    }

    // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
    // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)

    return dummy.next;

    }
}

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