leetcode143. Reorder List

medium程度题

题目:

Given a singly linked list L: L0→L1→…→Ln-1→Ln,

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,

Given 

{1,2,3,4}

, reorder it to 

{1,4,2,3}

.
我想的是把n个节点存入容器，然后遍历容器改造节点，时间复杂度O(N)，空间复杂度O（N）
AC解:

class Solution {
public:
void reorderList(ListNode* head)
{
if (head == nullptr || head->next == nullptr)
return ;
vector<ListNode*> vec;
//遍历一遍链表,所有节点存入vec
ListNode *cur = head;
while (cur)
{
vec.push_back(cur);
cur = cur->next;
}
//改造节点
auto first = vec.begin();
auto second = vec.end() - 1;
while (first < second)
{
(*first)->next = *second;
first++;
(*second)->next = *first;
second--;
}

(*first)->next = nullptr;
}
};

另有一种空间复杂度O(1)的解法，把链表拆分成两个，将后面的链表翻转，然后合并两个链表
AC解:

class Solution {
public:
void reorderList(ListNode* head)
{
if (head == nullptr || head->next == nullptr)
return ;
ListNode *fast = head,*slow = head,*prev = nullptr;
//拆分链表
while (fast && fast->next)
{
prev = slow;
slow = slow->next;
fast = fast->next->next;
}

prev->next = nullptr;
slow = reverse(slow);
//合并两个链表
ListNode *cur = head;
while (cur->next)
{
ListNode *temp = cur->next;
cur->next = slow;
slow = slow->next;
cur->next->next = temp;
cur = temp;
}

cur->next = slow;
}

ListNode* reverse(ListNode *head)
{
if (head == nullptr || head->next == nullptr)
return head;
ListNode *prev = head;
for (ListNode *cur = head->next,*next = cur->next; cur;)
{
head->next = next;
cur->next = prev;
prev = cur;
cur = next;

next = next ? next->next : nullptr;
}

return prev;
}
};

两种解法时间性能上没有差别