leetcode 92. 反转链表 II

#include"iostream"
using namespace std;
struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode* result = head;  //设置个用于输出结果的指针，=head
        ListNode* pre_head=NULL, * modify_list_tail=NULL, * new_head=NULL;//需要三个辅助的指针，一定要先初始化
        int len = n - m + 1;//求出要操作的节点个数，也就是在设置逆序的次数
        while (head != NULL && --m) {   //找到开始逆序的位置
            pre_head = head;            //pre_head是要中间逆序链表头的前一个节点
            head = head->next;       
        }
        modify_list_tail = head;       //该头结点是逆序完成后的尾结点
        new_head = NULL;
        while(head&&len) {             //逆序过程
            ListNode *next = head->next;
            head->next = new_head;
            new_head = head;
            head = next;
            len--;
        }
        modify_list_tail->next = head;  //原头节点变尾结点了，要将其next指向当前head
        //判断是不是从第一个节点开始的逆序
        if (pre_head) {                 //不是
            pre_head->next = new_head;
        }
        else                           //是
        {
            result = new_head;
        }
        return result;
    }
};


int main()
{
    int m, n;
    m = 2;
    n = 4;
    ListNode a(1);
    ListNode b(2);
    ListNode c(3);
    ListNode d(4);
    ListNode e(5);
    a.next = &b;
    b.next = &c;
    c.next = &d;
    d.next = &e;
    e.next = NULL;
    Solution s1;
    ListNode* head = &a;//遍历需要个辅助的头指针啊
    printf("before reverse:\n");
    while (head) {
        printf("%d\n", head->val);
        head = head->next;
    }
    printf("after reverse:\n");
    head = s1.reverseBetween(&a,m,n);
    while (head) {
        printf("%d\n", head->val);
        head = head->next;
    }
    system("pause");
    return 0;
}