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.
解题思路：
指针操作即可，旋转参考Java for LeetCode 025 Reverse Nodes in k-Group JAVA实现如下：
static public ListNode reverseBetween(ListNode head, int m, int n) {
ListNode result = new ListNode(0);
result.next = head;
if (m >= n || m <= 0)
return result.next;
head = result;
for (int i = 0; i < m - 1; i++)
head = head.next;
Stack<Integer> stk = new Stack<Integer>();
ListNode temp = head.next;
for (int i = 0; i <= n - m; i++)
if (temp != null) {
stk.push(temp.val);
temp = temp.next;
}
if (stk.size() == n - m + 1) {
while (!stk.isEmpty()) {
head.next = new ListNode(stk.pop());
head = head.next;
}
head.next = temp;
}
return result.next;
}
Java for LeetCode 092 Reverse Linked List II
原文：http://www.cnblogs.com/tonyluis/p/4516613.html