题解 | #链表内指定区间反转#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
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) {
        // C++ 时间O(n) 空间O(1)解法 拆下需反转部分，再拼接回去
        if(m==n) return head;

        ListNode* start = nullptr;
        ListNode* end = nullptr;
        ListNode* start_pre = nullptr;
        ListNode* end_next = nullptr;

        ListNode* cur = head;
        while(--m) {
            start_pre = cur;
            cur=cur->next;
        }
        start = cur;
       
        cur = head;
        while(--n) {
            cur=cur->next;
        }
        end = cur;
        end_next = cur->next;

        end->next = nullptr;
        cur = start;
        ListNode* pre = nullptr;
        while(cur!=nullptr) {
            ListNode* cur_next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = cur_next; 
        }


        if(start_pre) {
            start_pre->next = pre;
            pre = head;
        }

        cur = pre;
        while(cur->next) cur=cur->next;
        cur->next = end_next;
        return pre;
    }
};