原题连接：Leetcode 206. Reverse Linked List
Given the head of a singly linked list, reverse the list, and return the reversed list.
Example 1:
Input: head = [1,2,3,4,5] Output: [5,4,3,2,1]
Example 2:
Input: head = [1,2] Output: [2,1]
Example 3:
Input: head = [] Output: []
Constraints:
- The number of nodes in the list is the range [0, 5000].
- -5000 <= Node.val <= 5000
做法一：双指针 思路：
利用两个指针指向相邻的两个元素，实现局部反转，然后遍历整个数组，实现整体反转
c++代码：
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* pre = head;
ListNode* cur = NULL;
while(pre != NULL){
// 先记录pre->next
ListNode* tmp = pre->next;
pre->next = cur;
cur = pre;
pre = tmp;
}
return cur;
}
};
复杂度分析：
- 时间复杂度：O(n)，需要遍历整个数组
- 空间复杂度：O(1)，常数个临时变量
