[2074] Reverse Nodes in Even Length Groups

這個算是group reverse linked list 系列吧

雖然對於一次要反轉幾個有點卡住

想說要先算起來還是怎樣的，但先弄起來反而就知道要把reverse 塞在哪兒了吧。

整理過後的code 長這樣，我懶得把它弄的更漂亮了XD 先降紫～～～～

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* struct ListNode *next;

* };

*/

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* struct ListNode *next;

* };

*/

struct ListNode* reverseEvenLengthGroups(struct ListNode* head) {

struct ListNode *ptr = head;

int check = 0;

while (ptr != NULL)

{

ptr = ptr->next;

check++;

}

int move;

ptr=head;

struct ListNode *ptrPre= NULL;

struct ListNode *pre, *curr, *next;

for (int i=1; check >0; i++)

{

if (check - i < 0) //last group

{

if (check %2==0)

{

pre = NULL;

curr = ptr;

while (curr != NULL)

{

next = curr->next;

curr->next = pre;

pre = curr;

curr = next;

}

ptrPre->next = pre;

}

break;

}

check-= i;

move = i;

if (i%2==0)

{

pre = NULL;

curr = ptr;

while (move > 0)

{

next = curr->next;

curr->next = pre;

pre = curr;

curr = next;

move --;

}

ptrPre->next = pre;

ptr->next = curr;

ptrPre=ptr;

ptr = ptr->next;

}

else // move ptr

{

while (move>0)

{

move--;

ptrPre= ptr;

ptr = ptr->next;

}

}

}

return head;

}

下面是一點都不重要的，算有幾個要reverse，跟最後一個要不要reserve ，的code  XD

struct ListNode* reverseEvenLengthGroups(struct ListNode* head) {

struct ListNode *ptr = head;

int count = 0;

while (ptr != NULL)

{

ptr = ptr->next;

count++;

}

int check = count;

int evenCount=0;

bool last = false;

ptr=head;

for (int i=1; check >0; i++)

{

if (check - i < 0)

{

if (check %2==0)

{

last= true;

evenCount++;

}

break;

}

check-= i;

int move = i;

if (i%2==0)

evenCount++;

}

// printf("%d %d \n", count, evenCount);

return head;

}