Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You may not modify the values in the list's nodes, only nodes itself may be changed.
Example 1:
Given 1->2->3->4, reorder it to 1->4->2->3.
Example 2:
Given 1->2->3->4->5, reorder it to 1->5->2->4->3. Solution：
使用快慢指针，得到后半部分的链表
使用栈存储后半部分的链表来替代翻转链表
class Solution {
public:
void reorderList(ListNode* head) {
if (head == nullptr || head->next == nullptr)return;
ListNode* slow = head, *fast = head;
while (fast && fast->next)
{
slow = slow->next;
fast = fast->next->next;
}
fast = slow->next;
slow->next = nullptr;
stack<ListNode*>s;
while (fast)
{
s.push(fast);
fast = fast->next;
}
slow = head;
while (!s.empty())
{
ListNode* p = slow->next;
slow->next = s.top();
s.pop();
slow->next->next = p;
slow = p;
}
return;
}
};