第一个想法随着vector保存全部Node*
表拼接出来
void reorderList(ListNode *head) {
vector<ListNode*> content;
ListNode * cur = head;
while (cur)
{
content.push_back(cur);
cur = cur->next;
}
int size = content.size();
cur = NULL;
for (int i = 0; i <= size - 1 - i; i++)
{
if (cur) cur->next = content[i];
if (i != size - i - 1) content[i]->next = content[size - 1 - i];
cur = content[size - 1 - i];
}
if (cur) cur->next = NULL;
}
另外一种思路能够分成下面几个步骤:
1. 找到中间的节点nMid
2. 翻转nMid到末尾的链表
3. 拼接head和nMid
代码例如以下
void reorderList(ListNode *head) {
if (head == NULL) return;
ListNode * mid = head;
ListNode * endOfHead;
ListNode * end = head;
while (end)
{
end = end->next;
endOfHead = mid;
mid = mid->next;
if (end) end = end->next;
}
endOfHead->next = NULL;//end of first half
//reverse
if (mid == NULL) return;
ListNode* p0, * p1, * p2;
p1 = mid;
p2 = mid->next;
p1->next = NULL;
while (p2)
{
p0 = p1;
p1 = p2;
p2 = p2->next;
p1->next = p0;
}
mid = p1;
//concat
ListNode * newHead = head;
while (mid && head)
{
head = head->next;
newHead->next = mid;
newHead = newHead->next;
mid = mid->next;
newHead->next = head;
newHead = newHead->next;
}
}
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