LeetCode 445 Add Two Numbers II（栈+链表）

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 8 -> 0 -> 7

题目大意：给出两个非空链表，求两个链表所代表的数字之和。

解题思路：遍历每个链表，将值存入两个栈中，然后对栈顶元素操作即可。

代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */ struct Node{    int val;    struct Node* next;};struct StackNode{    int size;    struct Node* top;};typedef struct StackNode* Stack;void makeEmpty(Stack stk){    stk->size = 0;    stk->top = NULL;}Stack createStack(){    Stack stk = malloc(sizeof(struct StackNode));    makeEmpty(stk);    return stk;}void push(Stack stk, int x){    struct Node* tmp = malloc(sizeof(struct Node));    tmp->next = NULL;    tmp->val = x;    if(stk->size == 0){        stk->top = tmp;    }else{        tmp->next = stk->top;        stk->top = tmp;    }    stk->size++;}void pop(Stack stk){    struct Node* tmp = stk->top;    stk->top = tmp->next;    stk->size--;    free(tmp);}int top(Stack stk){    if(stk && stk->top)        return stk->top->val;    return 0;}void destroyStack(Stack stk){    while(stk->size){        pop(stk);    }    free(stk);}void traverse(struct ListNode* list, Stack stk){    while(list){        push(stk, list->val);        list = list->next;    }}struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {    Stack stk1 = createStack();    Stack stk2 = createStack();    Stack ans  = createStack();        traverse(l1, stk1);    traverse(l2, stk2);    int sum = 0;    while(stk1->size || stk2->size){        if(stk1->size){            sum += top(stk1);            pop(stk1);        }        if(stk2->size){            sum += top(stk2);            pop(stk2);        }        push(ans, sum % 10);        sum /= 10;    }    if(sum) push(ans, sum);    destroyStack(stk1);    destroyStack(stk2);        return ans->top;}