1.题目描述
Reverse a linked list from position m to n. Do it in one-pass.
反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL
2.Solutions
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public static ListNode (ListNode head, int m, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode cur1 = dummy;
ListNode pre1 = null;
for(int i=0;i<m;i++){
pre1 = cur1;
cur1 = cur1.next;
}
ListNode cur2 = cur1;
ListNode pre2 = pre1;
ListNode next;
for(int i=m;i<=n;i++){
next = cur2.next;
cur2.next = pre2;
pre2 = cur2;
cur2 = next;
}
pre1.next = pre2;
cur1.next = cur2;
return dummy.next;
}
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参考：【链表】打卡2：如何优雅着反转单链表
（完）
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