LeetCode 第92题 反转链表 II
作者:互联网
反转链表 II
反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。
说明:
1 ≤ m ≤ n ≤ 链表长度。
示例:
输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL
1 class Solution97 { 2 3 public ListNode reverseBetween(ListNode head, int m, int n) { 4 if (head == null || head.next == null || m >= n) { 5 return head; 6 } 7 8 ListNode dummyHead = new ListNode(0); 9 dummyHead.next = head; 10 ListNode pre = dummyHead; 11 ListNode left = dummyHead.next; 12 ListNode right = dummyHead.next; 13 14 for (int i = 0; i < n - m; i++) { 15 right = right.next; 16 } 17 for (int i = 0; i < m - 1; i++) { 18 pre = left; 19 left = left.next; 20 right = right.next; 21 } 22 for (int i = 0; i < n - m; i++) { 23 pre.next = left.next; 24 left.next = right.next; 25 right.next = left; 26 left = pre.next; 27 } 28 29 return dummyHead.next; 30 } 31 }
标签:II,right,ListNode,int,next,链表,dummyHead,92,left 来源: https://www.cnblogs.com/rainbow-/p/10348977.html