LeetCode之“链表”：Reverse Linked List && Reverse Linked List II

　 1. Reverse Linked List

　　题目链接

　　题目要求：

　　Reverse a singly linked list.

　　Hint:

　　A linked list can be reversed either iteratively or recursively. Could you implement both?

　　初想用递归来实现应该会挺好的，但最终运行时间有点久，达到72ms，虽然没超时。具体程序如下：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */

 ListNode* recurList(ListNode* p)
 {
     if (p->next != NULL)
     {
         ListNode* tmp = p->next;
         if (tmp->next == NULL)
         {
             tmp->next = p;
             p->next = NULL;
             return tmp;
         }
     }

     ListNode* ret = recurList(p->next);
     ListNode* tmp = ret;
     while (tmp->next != NULL)
         tmp = tmp->next;
     tmp->next = p;
     p->next = NULL;
     return ret;
 }

 ListNode* reverseList(ListNode* head) {
     if (head == NULL || head->next == NULL)
         return head;

     ListNode* tail = recurList(head);
     return tail;
 }

　　在LeetCode论坛发现了一个8ms的解法，不得不为其简洁高效合彩！具体程序如下：

 ListNode* reverseList(ListNode* head) {
     ListNode *curr = head, *prev = nullptr;
     while (curr) {
         auto next = curr->next;
         curr->next = prev;
         prev = curr, curr = next;
     }
     return prev;
 }

　 2. Reverse Linked List II

　　题目链接

　　题目要求：

　　Reverse a linked list from position m to n. Do it in-place and in one-pass.

　　For example:
　　Given 1->2->3->4->5->NULL, m = 2 and n = 4,

　　return 1->4->3->2->5->NULL.

　　Note:
　　Given m, n satisfy the following condition:
　　1 ≤ m ≤ n ≤ length of list.

　　该题的解法可以将一个链表分成三个部分，即1~m-1、m~n、n+1~最后一个元素3个部分，且仅对中间部分进行翻转，最后再将这3个部分合并。具体程序如下：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* reverseBetween(ListNode* head, int m, int n) {
         if(!head || !head->next || m >= n)
              return head;

         ListNode *start = head;
         ListNode *preList = nullptr, *postList = nullptr;

         int k = ;
         while(k < m)
         {
             preList = start;
             start = start->next;
             k++;
         }

         ListNode *prev = nullptr;
         while(k <= n)
         {
             if(k == n)
                 postList = start->next;
             auto next = start->next;
             start->next = prev;
             prev = start;
             start = next;
             k++;
         }

         ListNode *end = prev;
         while(end && end->next)
             end = end->next;

         if(preList)
             preList->next = prev;
         else
             head = prev;
         end->next = postList;

         return head;
     }
 };