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
     */
	// 翻转head->...->tail链表的函数
    pair<ListNode*, ListNode*> reverse0(ListNode* head, ListNode* tail){
        ListNode* prev = tail->next;
        ListNode* p = head;
        while (prev != tail) {
            ListNode* nex = p->next;
            p->next = prev;
            prev = p;
            p = nex;
        }
        return {tail, head};// 返回新链表的头和尾
    }

    ListNode* reverseBetween(ListNode* head, int m, int n) {
        // write code here
        int cnt = 1;
        ListNode* hair = new ListNode(-1);
        hair->next = head;
        ListNode* pre = hair;
        

        while(cnt++ < m){
            pre = pre->next;
        }
        ListNode* first = pre->next;
        ListNode* end = first;
        while(cnt++ <= n){
            end = end->next;
        }
        ListNode* next = end->next;
        pair<ListNode*, ListNode*> res = reverse0(first, end);
        // tie(first, end) = reverse0(first, end);
        first = res.first;
        end = res.second;
	  // 记得连接新的首尾
        pre->next = first;
        end->next = next;

        return hair->next;

    }
};