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LoeiJe

:D 获取中...

何以解忧?唯有暴富

leetcode-2

leetcode 2

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// 我写的....
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode *first_l1 = l1, *first_l2 = l2;
//ListNode *ans, *root;
ListNode *ans = new ListNode(0);
ListNode *root = ans;
int flag = 0; // 进位标记,初始0
int aux = 0;
int x, y;
while(first_l1 || first_l2) {
if(first_l1) x = first_l1->val;
else x = 0;
if(first_l2) y = first_l2->val;
else y = 0;
aux = x + y + flag;
flag = aux / 10; aux = aux % 10;

ListNode *temp = new ListNode(aux);
ans->next = temp;
ans = ans->next;

if(first_l1) first_l1 = first_l1->next;
if(first_l2) first_l2 = first_l2->next;
}
if(flag) {
ListNode *temp = new ListNode(1);
ans->next = temp;
}
return root->next;
}
};

// 别人写的
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode ans(0), *p = &ans;
int aux = 0, flag = 0;
while(l1 || l2 || flag) {
aux = 0;
if(l1) aux += l1->val;
if(l2) aux += l2->val;
aux += flag;
flag = aux / 10;
aux = aux % 10;

ListNode *new_node = l1 ? l1 : l2; // 不新建节点 使用原数据节点
if(nullptr == new_node) new_node = new ListNode(aux);

new_node->val = aux; //使用元数据节点时 重新赋值
p->next = new_node;
p = p->next;
if(l1) l1 = l1->next;
if(l2) l2 = l2->next;
}
return ans.next;
}
};