题解 | #链表内指定区间反转#

基础题，考察反转单链表。需要注意细节处理。
用了一个while遍历链表，常熟个变量存放数据。时间复杂度O(n)，空间复杂度O(1)。满足题目要求。

/**
 * 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) {
        // write code here
        if(n == 1 || m == n){
            return head;
        }
        ListNode* p = head;
        ListNode* before = nullptr;
        ListNode* after = nullptr;
        ListNode* pre = nullptr;
        ListNode* front = nullptr;
        ListNode* back = nullptr;      
        int i = 0;
        m--;
        n--;
        while(p){
            if(i == m - 1){
                before = p;
            }
            else if(i == n + 1){
                after = p;
            }
            if(i == m){
                front = p;
            }
            else if(i == n){
                back = p;
            }
            if(i > m && i <= n){
                ListNode* temp = p;
                p = p->next;
                temp->next = pre;
                pre = temp;
            }
            else
            {
                pre = p;
                p = p->next;
            }
            i++;
        }
        if(before){
            before->next = back;
        }
        else{
            head = back;
        }
        front->next = after;
        return head;
    }
};