[Lintcode]Reverse Linked List II 翻转链表 II

最新推荐文章于 2022-01-17 17:31:54 发布

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最新推荐文章于 2022-01-17 17:31:54 发布

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分类专栏：算法文章标签： lintcode

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本文介绍了一种使用双指针技术来实现链表中指定区间的反转。通过具体示例展示了如何将从位置m到n的链表部分进行反转，提供了一个清晰的实现思路和完整的Java代码实现。

Reverse a linked list from position m to n.

Example

Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL.

分析：实现题。双指针。

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param ListNode head is the head of the linked list 
     * @oaram m and n
     * @return: The head of the reversed ListNode
     */
    public ListNode reverseBetween(ListNode head, int m , int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode first = dummy, second = null;

        while(m > 1) {
            first = first.next;
            m--;
            n--;
        }
        
        second = first.next;
        first.next = null;
        ListNode secondHead = null, firstEnd = second;
        while(n > 0) {
            secondHead = second.next;
            ListNode tmp = first.next;
            first.next = second;
            second.next = tmp;
            second = secondHead;
            n--;
        }
        firstEnd.next = secondHead;
        return dummy.next;
    }
}