单链表的基本处理，涉及逆序，计算长度等操作。
题意要求：
1->2->3->4 => 1->4->2->3 间隔操作。
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
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class Solution {
public:
void reorderList(ListNode *head) {
int len = getLength(head);
if(len <= 2)return ;
int pos = len%2==0?len/2:len/2+1;
ListNode * headend = getListNode(head,pos-1);
ListNode * head2 = getListNode(head,pos);
headend -> next = NULL;
head2 = reverseListNode(head2);
ListNode * headc = head;
while(head2 != NULL) {
ListNode * headnext = head->next;
ListNode * head2next = head2->next;
head->next = head2;
head2->next = headnext;
head = headnext;
head2 = head2next;
}
head = headc;
}
private:
ListNode * reverseListNode(ListNode * head) {
ListNode * before = head;
ListNode * now = head->next;
while(now != NULL) {
ListNode * next = now->next;
now->next = before;
before = now;
now = next;
}
head->next = NULL;
return before;
}
int getLength(ListNode *head) {
int len = 0;
while(head != NULL) {
len++;
head = head->next;
}
return len;
}
ListNode *getListNode(ListNode *head,int pos) {
int i = 0;
while(i < pos) {
head = head->next;
i++;
}
return head;
}
};
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