Leetcode: Reorder List

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}

.

分析：注意题目的要求，(1)in-place (2)not altering the nodes' values。最简单的方式是brute force，先找到last node，然后插入到相应的位置，重复上述过程直到遇到原来的中间节点变为last node或者second last node。brute force的时间复杂度是O(n^2)。但此题有O(n)的解法，先找到中间节点，然后将原来的list断成前后两部分，对后半部分list进行reverse操作，再与前半部分list merge。O(n)解法代码如下：

class Solution {
public:
void reorderList(ListNode *head) {
if(head == NULL || head->next == NULL || head->next->next == NULL) return;

int len = 0;
for(ListNode *p = head; p; p = p->next)
len++;

ListNode *mid = head;
for(int i = 0; i < len/2; i++)
mid = mid->next;

ListNode *sec_head = reverse(mid->next);
mid->next = NULL;

merge(head, sec_head);
}

ListNode * reverse(ListNode *head){
if(head == NULL || head->next == NULL) return head;

ListNode *dummy = new ListNode(-1);
dummy->next = head;

for(ListNode *prev = head, *p = head->next; p; p = prev->next){
prev->next = p->next;
p->next = dummy->next;
dummy->next = p;
}

return dummy->next;
}

void merge(ListNode *head1, ListNode *head2){
while(head1 && head2){
ListNode *tmp1 = head1->next;
ListNode *tmp2 = head2->next;
head1->next = head2;
head2->next = tmp1;
head1 = tmp1;
head2 = tmp2;
}
}
};