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.]
代码语言：javascript复制 public ListNode reverseBetween(ListNode head, int m, int n) {
if (head == null || head.next == null || m == n) {
return head;
}
ListNode fakeHead = new ListNode(-1);
fakeHead.next = head;
// 先向后移m步
ListNode pre = fakeHead;
for (int i = 1; i < m; i ) {
pre = pre.next;
}
// 对后面的n-m个结点，逆置
ListNode mNode = pre.next;
for (int i = m; i < n; i ) {
ListNode cur = mNode.next;
mNode.next = cur.next;
cur.next = pre.next;
pre.next = cur;
}
return fakeHead.next;
}
