[725] Split Linked List in Parts （TBD）

覺得微瑣碎的題目？！（thinking 圖）

雖然有點想要一個迴圈處理，但感覺～～～分成不同size兩次好像也不錯？！XD

因為中斷了所以就先降子吧～

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* struct ListNode *next;

* };

*/

/**

* Note: The returned array must be malloced, assume caller calls free().

*/

struct ListNode** splitListToParts(struct ListNode* head, int k, int* returnSize){

int total = 0;

struct ListNode *ptr = head;

while (ptr != NULL)

{

total++;

ptr = ptr->next;

}

int remain = total;

int space = total / k;

int extra = total%k; //number of (space+1)

*returnSize = k ;

struct ListNode** ret = calloc ((*returnSize), sizeof (struct ListNode*));

int idx=0;

struct ListNode *pre=NULL;

while (extra > 0)

{

ret[idx] = calloc (space +1, sizeof (struct ListNode));

int spacecount = space+1;

ptr=head;

while (spacecount> 0 && ptr != NULL)

{

pre = ptr;

ptr = ptr->next;

spacecount--;

}

pre->next = NULL;

ret[idx] =head;

head = ptr;

extra--;

remain -= (space+1);

idx++;

}

while (remain > 0)

{

ret[idx] = calloc (space, sizeof (struct ListNode));

int spacecount = space;

ptr=head;

while (spacecount> 0 && ptr != NULL)

{

pre = ptr;

ptr = ptr->next;

spacecount--;

}

pre->next = NULL;

ret[idx] =head;

head = ptr;

remain -= (space);

idx++;

}

for (;idx<(*returnSize);idx++)

ret[idx] =NULL;

return ret;

}