002. Add Two Numbers

002. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order 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.

Example:

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Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Solution

核心思想:正确处理进位
我们使用一个变量carry(初始化为0)来跟踪进位,遍历l1,l2的节点从低位到高位依次进行加法模拟,每位的计算完成后将carry带入下一次迭代。

Notes:carry的值必定为0或1,因为个位数相加(考虑进位)可能出现的最大和为:9+9+1=19。处理max(m, n)位时,要考虑是否有额外进位。

特殊情况:

测试用例 说明
l1 = [0,1], l2 = [0,1,2] 当一个列表比另一个列表长时
l1 = [], l2 = [0,1,2] 当其中一个列表为空时
l1 = [1], l2 = [9,9,9] 求和运算最后可能出现额外的进位,这一点很容易被遗忘

Complexity

时间复杂度:O(max(m, n)),假设l1和l2的长度为m和n,对链表的处理最多执行max(m, n)次;
时间复杂度:O(max(m, n)),新链表的节点数最多为max(m, n) + 1个。

Code

初始版本:

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class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* res = new ListNode(0);
ListNode* cur = res;

int carry = 0;
while (l1 != nullptr && l2 != nullptr) {
cur->next = new ListNode((l1->val + l2->val + carry) % 10);
carry = (l1->val + l2->val + carry) / 10;
cur = cur->next;
l1 = l1->next; l2 = l2->next;
}

while (l1 != nullptr) {
cur->next = new ListNode((l1->val + carry) % 10);
carry = (l1->val + carry) / 10;
cur = cur->next;
l1 = l1->next;
}

while (l2 != nullptr) {
cur->next = new ListNode((l2->val + carry) % 10);
carry = (l2->val + carry) / 10;
cur = cur->next;
l2 = l2->next;
}

if (carry != 0)
cur->next = new ListNode(carry);

return res->next;
}
};

简化版本:

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class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode preHead(0), *p = &preHead;
int extra = 0;
while (l1 || l2 || extra) {
if (l1) extra += l1->val, l1 = l1->next;
if (l2) extra += l2->val, l2 = l2->next;
p->next = new ListNode(extra % 10);
extra /= 10;
p = p->next;
}
return preHead.next;
}
};

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