LeetCode 92
https://leetcode.cn/problems/reverse-linked-list-ii/description/
难度：中等
这道题需要翻转中间一段链表。我们定义这段链表的前一个节点为 p0。翻转结束之后，需要修改 p0->next->next = cur, p0->next = pre。
时间复杂度：O(right)。
空间复杂度：O(1)。
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int left, int right) {
ListNode* dummy = new ListNode();
dummy->next = head;
ListNode* p0 = dummy;
for (int i = 0; i < left - 1; i++) {
p0 = p0->next;
}
ListNode* cur = p0->next;
ListNode* pre = nullptr;
for (int i = 0; i < right - left + 1; i++) {
ListNode* nxt = cur->next;
cur->next = pre;
pre = cur;
cur = nxt;
}
p0->next->next = cur;
p0->next = pre;
return dummy->next;
}
};