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}.
思路：本题未限制额外的space，所以可先得到该转换的节点，将该节点之后的节点存储在栈中，即可实现之后节点的反转。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void reorderList(ListNode *head) {
stack<ListNode*> S;
ListNode* ptr = head;
ListNode* top;
ListNode* previous;
int n = 0;
while(ptr != NULL)
{
++n;
ptr = ptr->next;
}
if (n == 1 || n == 2 || n == 0)
{
return;
}
n = n/2 + n%2;
ptr = head;
while(n--)
{
ptr = ptr->next;
}
while(ptr != NULL)
{
S.push(ptr);
ptr = ptr->next;
}
previous = head;
while(!S.empty())
{
top = S.top();
S.pop();
ptr = previous->next;
previous->next = top;
top->next = ptr;
previous = ptr;
}
previous->next = NULL;
}
};