[2181] Merge Nodes in Between Zeros

寫出了好可怕的東西 (嘆氣)

想要把多出來的node free 掉，感覺是自討苦吃(?)

到底面試的時候在不在意這個呢QQ

把別人美美的code 貼在下面，過幾天再回來寫一次QQ

ps. 原來可以在while 裡面再搞一個while, 一次將中間的val 相加到底。
別人為什麼就那麼聰明呢QQ

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* struct ListNode *next;

* };

*/

#if 1

struct ListNode* mergeNodes(struct ListNode* head) {

struct ListNode* p1 = head;

struct ListNode* p2 = head;

int num = 0;

while (p2->next != NULL)

{

if (p2->next->val == 0)

{

p1->val = num;

num = 0;

struct ListNode* p_next = p1->next;

while (p_next!= p2->next)

{

struct ListNode* p_free;

p_free = p_next;

p_next = p_next->next;

p_free = NULL;

free(p_free);

}

if (p_next->next == NULL)

{

p1->next = NULL;

return head;

}

p1->next = p_next;

p1= p_next;

}

else

num += p2->next->val;

p2 = p2->next;

}

return head;

}

#else

struct ListNode* mergeNodes(struct ListNode* head) {

struct ListNode* t1 = head->next;

struct ListNode* t2 = t1;

while(t2!=NULL){

int sum = 0;

while(t2->val != 0){

sum+=t2->val;

t2=t2->next;

}

t1->val = sum;

t2=t2->next;

t1->next = t2;

t1=t1->next;

}

return head->next; // skip first 0

}

#endif

（隔兩天更）

把它塞到前面的版本: 

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* struct ListNode *next;

* };

*/

struct ListNode* mergeNodes(struct ListNode* head) {

struct ListNode* ptr = head->next;

struct ListNode* p_sum = head;

int num = 0;

while (ptr != NULL)

{

//printf("%d \n", ptr->val);

if (ptr->val != 0)

{

num +=ptr->val;

ptr = ptr->next;

continue;

}

p_sum->val = num;

num = 0;

if (ptr->next == NULL)

{

p_sum->next= NULL;

break;

}

p_sum->next = ptr;

p_sum = ptr;

ptr = ptr->next;

}

return head;

}

把它塞到後面的版本:

struct ListNode* mergeNodes(struct ListNode* head) {

struct ListNode* ptr=head->next;

struct ListNode* p_sum=head;

int num = 0;

while (ptr!= NULL)

{

if(ptr->val !=0)

{

num += ptr->val;

ptr = ptr->next;

continue;

}

ptr->val = num;

p_sum->next = ptr;

p_sum= ptr;

ptr = ptr->next;

num = 0;

}

return head->next;

}