【Leetcode】【Easy】Reverse Linked List

题目：

Reverse a singly linked list.

解题:

反转单链表，不再多介绍了.

如果会“先条件->定参数->确定不变式->验证后条件”的思维方法，一定会bug free.

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

         ListNode* newhead = head;
         ListNode* curNode = NULL;
         head = head->next;
         newhead->next = NULL;

         while (head) {
             curNode = head;
             head = head->next;
             curNode->next = newhead;
             newhead = curNode;
         }

         return newhead;
     }
 };

写完初始版本后，再考虑如何去除循环前冗余的赋值，洁简代码：

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

         ListNode* newhead = NULL;
         ListNode* nextNode = NULL;

         while (head) {
             nextNode = head->next;
             head->next = newhead;
             newhead = head;
             head = nextNode;
         }

         return newhead;
     }
 };