反转链表 II

题解：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseBetween(struct ListNode* head, int left, int right) {
   struct ListNode* dummy=(struct ListNode*)malloc(sizeof(struct ListNode));
   dummy->next=head;
   struct ListNode *newHead,*temp,*steady=head,*last=dummy;
   for(int i=0;i<left-1;i++){
    steady=steady->next;
    last=last->next;
   }
   newHead=steady;
   temp=steady->next;
   for(int i=0;i<right-left;i++){
    steady->next=temp->next;
    temp->next=newHead;
    newHead=temp;
    temp=steady->next;
   }
   last->next=newHead;
   return dummy->next;
}

结果：时间复杂度O(N),空间复杂度O(1)