[445] Add Two Numbers II

竟然是這題！！！！！！

幸好我還有想到reverse 這件事Orz

但reverse 完卻忘記reverse 回來真是太悲了 Orz

而居然leetcode 上面是要求不要對input linked list 做reverse !

看很多人都提到stack，總覺得這跟reverse沒什麼不一樣啊哈哈哈哈哈XD!

總之那就胡亂用個暴力法解決好了，把數字倒出來到array[100]裡面，讓它們去各字相加，最後存到array3[100]裡面，處理十進位問題，然後再建linked list ! 先降XD!

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* struct ListNode *next;

* };

*/

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {

int *stack1= calloc (100, sizeof(int));//0-99

int *stack2= calloc (100, sizeof(int));

int count1=0, count2=0;

struct ListNode *p1 = l1;

struct ListNode *p2 = l2;

while (p1 != NULL)

{

stack1[count1++]= p1->val;

p1= p1->next;

}

while (p2 != NULL)

{

stack2[count2++]= p2->val;

p2= p2->next;

}

int *stack3= calloc (100, sizeof(int));

int count3=0;

if (count1 >= count2)

{

int i=count1, j=count2;

while (j>0)

{

stack3[count3++]= stack1[--i]+stack2[--j];

}

while (i>0)

stack3[count3++]= stack1[--i];

}

else if (count1 < count2)

{

int i=count1, j=count2;

while (i >0)

{

stack3[count3++]= stack1[--i]+stack2[--j];

}

while (j>0)

stack3[count3++]= stack2[--j];

}

for (int i=0; i<count3; i++)

{

if (stack3[i]>=10)

{

stack3[i]-=10;

if (i+1 == count3)

count3++;

stack3[i+1]++;

}

}

struct ListNode *newHead= NULL;

struct ListNode *ptr;

newHead = calloc(1, sizeof(struct ListNode));

ptr = newHead;

while (count3>0)

{

ptr->val = stack3[--count3];

if (count3 >0)

ptr->next = calloc(1, sizeof(struct ListNode));

ptr = ptr->next;

}

return newHead;

}