题解 | #链表内指定区间反转#

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param m int整型 
     * @param n int整型 
     * @return ListNode类
     */
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode *dummy=nullptr;
        ListNode *h = head;
        for(int i=1; i<m; i++) {
            dummy = head;
            head = head->next;
        }
        ListNode *start = head;
        int revLen = n-m;
        for(int i=0; i<revLen; i++) {
            head = head->next;
        }
        ListNode *end = head->next;
        head->next = nullptr;
        ListNode *newHead = reverse(start);
        
        if(dummy == nullptr) {
            h = newHead;
        } else {
            dummy->next = newHead;
        }
        head = newHead;
        while(head->next != nullptr) {
            head = head->next;
        }
        head->next = end;
        return h;
    }

    ListNode* reverse(ListNode* head) {
        if(head == nullptr)
            return head;
        ListNode* dummy = head;
        ListNode* left = head;
        ListNode* right = head->next;
        while(right != nullptr) {
            ListNode* next = right->next;
            right->next = left;
            left = right;
            right = next;
        }
        dummy->next = nullptr;
        return left;
    }
    
};

借用上一题反转列表的逻辑，先找到要反转的区间，并标记好区间边界的链表节点，反转结束之后再接上即可。注意边界条件的处理