题解 | #链表内指定区间反转#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 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
 */
 typedef struct ListNode listnode;
 listnode*Buynode(int x)//获取新创建的链表节点空间
 {
    listnode*newnode = (listnode*)malloc(sizeof(listnode));
    newnode->val = x;
    newnode->next = NULL;
    return newnode;
 }
 void ListnodePop(listnode**phead,int x)//将数据头插到新链表中
 {
    listnode*newnode = Buynode(x);
    newnode->next = *phead;
    *phead = newnode;
 }
struct ListNode* reverseBetween(struct ListNode* head, int m, int n ) {
    if(m==n)
    {
        return head;
    }
    listnode*l1,*l2,*l3;
    l1 = head;
    int count = 0;
    listnode*st = NULL;
    while(l1)
    {
        ++count;
        if(count==n)//获取n位置的结点地址
        {
            if(l1->next!=NULL)
            l3 = l1->next;
            else if(l1->next ==NULL)
            l3 = l1;
        }
        if(count>=m &&count <=n)//循环头插
        {
            ListnodePop(&st,l1->val);
        }
         if(count==m-1)
        {
            l2 = l1;//获取m结点的前驱结点的地址
        }
        l1 = l1->next;
    }
    listnode*ptail = st;
    while(ptail->next)
    {
        ptail = ptail->next;
    }
    if(count>=n &&m==1)
     {
        if(count==n)
       {
            return st;
       }
        ptail->next = l3;
        return st;
     }
    else if(count>=n &&m>1)
    {
        l2->next = st;
        if(count==n)
        {
            return head;
        }
        ptail->next = l3;
        return head;
    }
    return head;
}