【重点】【链表】92. 反转链表 II

题目

法1：迭代写法

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Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
        dummy = ListNode(next=head)
        preLeftNode, leftNode = dummy, None  # left的前续节点,left对应节点
        for _ in range(left-1):
            preLeftNode = preLeftNode.next
        leftNode = preLeftNode.next

        pre, cur = None, leftNode
        for _ in range(right-left+1): # 共计right-left+1个结点需要反转
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt

        preLeftNode.next = pre
        leftNode.next = cur

        return dummy.next

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode preStart = dummy, postEnd = dummy; // 记录要反转链表串的前驱, 后继结点
        while (left > 1) {
            preStart = preStart.next;
            --left;
        }
        while (right >= 0) {
            postEnd = postEnd.next;
            --right;
        }

        // 翻转[preStart.next, postEnd)之间的结点
        ListNode pre = null, cur = preStart.next;
        while (cur != postEnd) {
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        preStart.next = pre; // 前段连接翻转后的起点
        cur = pre;
        while (cur.next != null) {
            cur = cur.next;
        }
        cur.next = postEnd; // 翻转后的终点连接后段

        return dummy.next;
    }
}