题解 | #链表内指定区间反转#

m=1的时候，必须要返回那个p2

 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */
/**
 * 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
 */
struct ListNode* reverseBetween(struct ListNode* head, int m, int n ) {
    // write code here
    struct ListNode* p0;
    struct ListNode* p1;
    struct ListNode* p2;
    struct ListNode* pn;
    p0 = head;
    p1 = head;
    p2 = head;
    pn = NULL;
    if(head == NULL || head->next ==NULL || (m==n))
    {
        return head;
    }
    else
    {
        for(int i=1;i<m;i++)
        {
            p0 = p1;
            p1 = p1->next;
            pn = p1;
        }
        for(int j=1;j<n;j++)
        {
            p2 = p2->next;
        }
        struct ListNode* cur;
        struct ListNode* pre;
        struct ListNode* nex;
        pre = p1;
        cur = pre->next;
        nex = NULL;
        while(pre != p2)  //反转
        {
            nex = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nex;
        }
        p0->next = pre;
        p1->next = cur;    //首尾相连
    }
    if (m==1)
    {
        return p2;
    }
    return head;
}