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| 1 | +/* |
| 2 | + * @lc app=leetcode.cn id=92 lang=java |
| 3 | + * |
| 4 | + * [92] 反转链表 II |
| 5 | + * |
| 6 | + * https://leetcode-cn.com/problems/reverse-linked-list-ii/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Medium (54.28%) |
| 10 | + * Likes: 1018 |
| 11 | + * Dislikes: 0 |
| 12 | + * Total Accepted: 205K |
| 13 | + * Total Submissions: 374.3K |
| 14 | + * Testcase Example: '[1,2,3,4,5]\n2\n4' |
| 15 | + * |
| 16 | + * 给你单链表的头指针 head 和两个整数 left 和 right ,其中 left 。请你反转从位置 left 到位置 right 的链表节点,返回 |
| 17 | + * 反转后的链表 。 |
| 18 | + * |
| 19 | + * |
| 20 | + * 示例 1: |
| 21 | + * |
| 22 | + * |
| 23 | + * 输入:head = [1,2,3,4,5], left = 2, right = 4 |
| 24 | + * 输出:[1,4,3,2,5] |
| 25 | + * |
| 26 | + * |
| 27 | + * 示例 2: |
| 28 | + * |
| 29 | + * |
| 30 | + * 输入:head = [5], left = 1, right = 1 |
| 31 | + * 输出:[5] |
| 32 | + * |
| 33 | + * |
| 34 | + * |
| 35 | + * |
| 36 | + * 提示: |
| 37 | + * |
| 38 | + * |
| 39 | + * 链表中节点数目为 n |
| 40 | + * 1 |
| 41 | + * -500 |
| 42 | + * 1 |
| 43 | + * |
| 44 | + * |
| 45 | + * |
| 46 | + * |
| 47 | + * 进阶: 你可以使用一趟扫描完成反转吗? |
| 48 | + * |
| 49 | + */ |
| 50 | + |
| 51 | +// @lc code=start |
| 52 | +/** |
| 53 | + * Definition for singly-linked list. |
| 54 | + * public class ListNode { |
| 55 | + * int val; |
| 56 | + * ListNode next; |
| 57 | + * ListNode() {} |
| 58 | + * ListNode(int val) { this.val = val; } |
| 59 | + * ListNode(int val, ListNode next) { this.val = val; this.next = next; } |
| 60 | + * } |
| 61 | + */ |
| 62 | +class Solution { |
| 63 | + public ListNode reverseBetween(ListNode head, int left, int right) { |
| 64 | + ListNode dummy = new ListNode(0); |
| 65 | + dummy.next = head; |
| 66 | + |
| 67 | + ListNode preLeftNode = dummy; |
| 68 | + for (int i = 1; i < left; i++) { |
| 69 | + preLeftNode = preLeftNode.next; |
| 70 | + } |
| 71 | + ListNode leftNode = preLeftNode.next; |
| 72 | + |
| 73 | + ListNode prev = leftNode; |
| 74 | + ListNode curr = leftNode.next; |
| 75 | + for (int i = left; i < right; i++) { |
| 76 | + ListNode next = curr.next; |
| 77 | + curr.next = prev; |
| 78 | + prev = curr; |
| 79 | + curr = next; |
| 80 | + } |
| 81 | + |
| 82 | + leftNode.next = curr; |
| 83 | + preLeftNode.next = prev; |
| 84 | + |
| 85 | + return dummy.next; |
| 86 | + } |
| 87 | +} |
| 88 | +// @lc code=end |
| 89 | + |
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