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.
public class Solution { /** * The algorithm is straightforward. We first move one pointer to the m-th node of * the list and then reverse the links of the following n - m nodes; * @param head --ListNode, the head node of a linked list * @param m --int, the start nodes * @param n --int, the ending nodes * @return ListNode --head of the new linked list * @author Averill Zheng * @version 2014-06-12 * @since JDK 1.7 */ public ListNode reverseBetween(ListNode head, int m, int n) { ListNode newHead = new ListNode(0); newHead.next = head; ListNode previous = null;; ListNode firstNode = newHead; int i = 0; while(i < m){ previous = firstNode; firstNode = firstNode.next; ++i; } //remember previous.next = firstNode; ListNode current = firstNode.next; ListNode tail = null; int j = 0; while(j < n - m){ tail = firstNode; firstNode = current; current = current.next; firstNode.next = tail; ++j; } previous.next.next = current; previous.next = firstNode; newHead = newHead.next; return newHead; } }
leetcode--Reverse Linked List II,布布扣,bubuko.com
leetcode--Reverse Linked List II
原文:http://www.cnblogs.com/averillzheng/p/3787759.html