【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;
    }
};