题解 | #反转链表#

这个方法简直太妙了，记录以下

假设： pre->cur->cur_next->pp

核心代码即：

cur_next=cur->next; //记录求next

cur->next=cur->next->next; //翻转cur和cur_next

cur_next->next=pre->next;

pre->next=cur_next;

/**
 * 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
        ListNode* dummyNode = new ListNode(-1);
        dummyNode->next =head;
        ListNode* pre = dummyNode;
        for(int i=0;i<m-1;i++){ //pre指向了第m-1个结点
            pre = pre->next;
        }
 
        ListNode* cur = pre->next; //开始翻转
        ListNode* Cur_next=nullptr ;
        for(int i=0;i<n-m;i++){
            Cur_next = cur->next;
            cur->next = Cur_next->next;
            Cur_next->next = pre->next;
            pre->next = Cur_next ;
        }
        return dummyNode->next;
    }
};