002 两数相加
题目
Tags Companiesadobe
| airbnb
| amazon
| bloomberg
| microsoft
给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。
请你将两个数相加,并以相同形式返回一个表示和的链表。
你可以假设除了数字 0 之外,这两个数都不会以 0 开头。
示例 1:
输入:l1 = [2,4,3], l2 = [5,6,4]
输出:[7,0,8]
解释:342 + 465 = 807.
示例 2:
输入:l1 = [0], l2 = [0]
输出:[0]
示例 3:
输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出:[8,9,9,9,0,0,0,1]
提示:
- 每个链表中的节点数在范围
[1, 100]
内 0 <= Node.val <= 9
- 题目数据保证列表表示的数字不含前导零
解决方案一
遵循加法法则,设置进位项,逐位相加。
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* head = new ListNode(), *p1 = l1, *p2 = l2, *p = head;
int addition = 0;
int sum = 0;
while(p1 && p2){
sum = p1->val + p2->val + addition;
addition = sum/10;
p->next = new ListNode(sum % 10);
p = p->next;
p1 = p1->next;
p2 = p2->next;
}
while(p1){
sum = p1->val + addition;
addition = sum/10;
p->next = new ListNode(sum % 10);
p = p->next;
p1 = p1->next;
}
while(p2){
sum = p2->val + addition;
addition = sum/10;
p->next = new ListNode(sum % 10);
p = p->next;
p2 = p2->next;
}
if(addition) p->next = new ListNode(addition);
return head->next;
}
};
效果:
1568/1568 cases passed (32 ms)
Your runtime beats 89.01 % of cpp submissions
Your memory usage beats 85.85 % of cpp submissions (69.4 MB)
解决方案二
利用递归将对应位数相加,在加上进位项。
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == NULL and l2 == NULL) return NULL;
else if (l1 == NULL) return l2;
else if (l2 == NULL) return l1;
int a = l1->val + l2->val;
ListNode *p = new ListNode(a % 10);
p->next = addTwoNumbers(l1->next,l2->next);
if (a >= 10) p->next = addTwoNumbers(p->next, new ListNode(1));
return p;
}
};
效果:
1568/1568 cases passed (56 ms)
Your runtime beats 17.92 % of cpp submissions
Your memory usage beats 5.19 % of cpp submissions (70.3 MB)