leetcode(92) Reverse Linked List II

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原题链接：https://leetcode.com/problems/reverse-linked-list-ii/

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.

将链表中从m到n位置倒换并输出新链表

反转链表的题一般有两种解法，图解见http://www.2cto.com/kf/201110/106607.html

使用第二种方法的java代码：

package leetcode;

public class ReverseLinkedListII {

	//https://leetcode.com/problems/reverse-linked-list-ii/
	public class ListNode {
		int val;
		ListNode next;

		ListNode(int x) {
			val = x;
		}
	}

	public ListNode reverseBetween(ListNode head, int m, int n) {

		ListNode root = new ListNode(0);
		ListNode p = root;
		root.next = head;

		for (int i = 1; i < m && p != null; i++) {
			p = p.next;
		}

		if (p != null) {
			ListNode q = p.next;
			ListNode r;

			int changeNumber = n - m + 1; // 交换节点的数目
			for (int i = 1; i < changeNumber && q.next != null; i++) {
				r = q.next;

				q.next = r.next;
				r.next = p.next;
				p.next = r;	//p的next永远是遍历到的最后一个
			}

			head = root.next;
		}

		return head;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
	}

}