【LeetCode & 剑指offer刷题】链表题1：14 反转链表 206. Reverse Linked List（系列）
【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
206. Reverse Linked List
Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both?
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
//迭代法
//O(n),O(1)
class Solution
{
public:
ListNode* reverseList(ListNode* head)
{
if(head == nullptr) return nullptr;
ListNode *pre = nullptr, *cur = head,*next; //三个指针分别保存之前，当前，下一个结点
while(cur)
{
next = cur->next; //保存原链表该结点的下一个结点，以免变换指向之后无法遍历到下一个结点
cur->next = pre; //变换指向
pre = cur; //更新指针
cur = next; //更新指针
}
return pre; //最后pre指向最后一个结点，cur指向null
}
};
//递归法(不太好理解)
/*
O(n) O(n)（来自递归栈）
Assume from node nk+1 to nm had been reversed and you are at node nk.
n1 → … → nk-1 → nk → nk+1 ← … ← nm
We want nk+1’s next node to point to nk.
So,
nk.next.next = nk;
*/
class Solution
{
public:
ListNode* reverseList(ListNode* head)
{
if(head == nullptr || head->next == nullptr) return head;
ListNode* p = reverseList(head->next); //递归之后为从后往前开始转向
head->next->next = head;
head->next = nullptr;
return p;
}
};
92. Reverse Linked List II
Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
/*
问题：翻转链表2（位置m~n翻转）
方法：对m~n位置处的链表进行翻转，然后连接m-1位置与n位置，m位置与n+1位置
例：
Input: 1 -> 2 -> 3 -> 4 -> 5 -> NULL, m = 2, n = 4
翻转得，1 <-> 2(m) <- 3 <- 4(n) 5 -> NULL
连接得，1 -> 4 -> 3 -> 2 -> 5 -> NULL
*/
class Solution
{
public:
ListNode* reverseBetween(ListNode* head, int m, int n)
{
//异常情况处理（严格来说，还需判断m,n值是否超出链表长度,题目已经有此条件）
if(!head || !head->next || m<0 || n<0 || m>=n) return head;
//prehead的应用：头结点前的结点，方便索引，方便分析（链表用1开头+prehead分析较好，序号可以与步数对应起来）
//且指向了链表的头结点，当m=1时，原头结点被换到后面，若返回head会出错，
//而用prehead会指向新头结点，故返回prehead.next不会出错
ListNode prehead(0);
prehead.next = head;
ListNode* cur = &prehead;
for(int i = 1; i <= m-1; i++)
cur = cur->next; //最后cur处在m-1位置
ListNode* pre = cur;
ListNode* next;
cur = cur->next; //让cur处在m位置
ListNode* prefirst = pre, *first = cur; //保存m-1位置和m位置处的结点
for(int i = m; i<=n; i++) //翻转位置m~n处的链表
{
next = cur->next;
cur->next = pre;
pre = cur; //更新指针
cur = next; //更新指针
}//退出时，pre在n处，cur在n+1处
prefirst->next = pre; //连接m-1与n位置
first->next = cur; //连接m与n+1位置
return prehead.next;
}
};
