题解 | #链表内指定区间反转#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @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 = new ListNode(-1);//在前面多增加一个-1，方便左段遍历
        dummy->next = head;//链接head链表
        ListNode* pNode = dummy;
        //获取左段尾节点，中段头节点，并将左段尾节点指向nullptr
        ListNode* ftail = pNode,*rhead = NULL, *rtail = NULL, *res = NULL;
        for (int i = 0; i < m - 1; i++)
            ftail = ftail->next;
        rhead = ftail->next;
        ftail->next = NULL;
        //对中段进行翻转指向，使用迭代的方式进行，一共迭代n-m+1次
        ListNode* prev = NULL, *pcur = rhead, *pnext = rhead->next;
        for (int i = m; i <= n;i++) {
            pcur->next = prev;
            prev = pcur;
            pcur = pnext;
            pnext = pnext->next;
        }
        //右段头节点赋给res
        res = pcur;
        //左段尾节点指向翻转后中段头节点
        ftail->next = prev;
        //翻转前中段头节点指向res
        rhead->next = res;
        //切除初始加的节点，删除，避免内存泄露
        ListNode* phead = dummy->next;
        dummy->next = NULL;
        delete dummy;
        return phead;
    }
};