2.两数相加(Add Two Numbers)M

英文题目

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 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:
```
Input: l1 = [0], l2 = [0]
Output: [0]
```

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Constraints:
The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.

中文题目

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

模拟

类似字符串相加的方法

时间复杂度 O(max(m,n))，空间复杂度O(1)

# python: 时间 60 ms, 击败 68.77%; 内存 16 MB, 击败 51.65%
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
from typing import Optional
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode(-1)
        cur = dummy
        carry = 0
        while l1 or l2:
            n1, n2 = 0, 0
            if l1:
                n1 = l1.val
                l1 = l1.next
            if l2:
                n2 = l2.val
                l2 = l2.next
            tmp = n1 + n2 + carry
            carry = tmp // 10
            cur.next = ListNode(tmp % 10)
            cur = cur.next
        if carry > 0:
            cur.next = ListNode(carry)
        return dummy.next

// c++: 时间 20 ms, 击败 91.91%; 内存 69.6 MB, 击败 89.18%
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode(-1);
        ListNode* cur = dummy;
        int carry = 0;
        while (l1 != nullptr || l2 != nullptr) {
            int n1 = 0, n2 = 0;
            if (l1 != nullptr) {
                n1 = l1->val;
                l1 = l1->next;
            }
            if (l2 != nullptr) {
                n2 = l2->val;
                l2 = l2->next;
            }
            int tmp = n1 + n2 + carry;
            cur->next = new ListNode(tmp % 10);
            carry = tmp / 10;
            cur = cur->next;
        }
        if (carry > 0) {
            cur->next = new ListNode(carry);
        }
        return dummy->next;
    }
};

// java: 时间 1 ms, 击败 100%; 内存 41.9 MB, 击败 48.53%
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        int carry = 0;
        while (l1 != null || l2 != null) {
            int n1 = 0, n2 = 0;
            if (l1 != null) {
                n1 = l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                n2 = l2.val;
                l2 = l2.next;
            }
            int tmp = n1 + n2 + carry;
            cur.next = new ListNode(tmp % 10);
            cur = cur.next;
            carry = tmp / 10;
        }
        if (carry > 0) {
            cur.next = new ListNode(carry);
        }
        return dummy.next;
    }
}

// go: 时间 16 ms, 击败 12.86%; 内存 4.3 MB, 击败 89.43%p
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    dummy := &ListNode{Val: -1}
    cur := dummy
    carry := 0
    for l1 != nil || l2 != nil {
        n1, n2 := 0, 0
        if l1 != nil {
            n1 = l1.Val
            l1 = l1.Next
        }
        if l2 != nil {
            n2 = l2.Val
            l2 = l2.Next
        }
        tmp := n1 + n2 + carry
        cur.Next = &ListNode{Val: tmp % 10}
        cur = cur.Next
        carry = tmp / 10
    }
    if carry > 0 {
        cur.Next = &ListNode{Val: carry}
    }
    return dummy.Next
}

// javascript: 时间 104 ms, 击败 46.23%; 内存 45.9 MB, 击败 61.39%
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
    let dummy = new ListNode(-1);
    let cur = dummy;
    let carry = 0;
    while (l1 || l2) { // 注意判空的方法
        let n1 = 0, n2 = 0;
        if (l1) {
            n1 = l1.val;
            l1 = l1.next;
        }
        if (l2) {
            n2 = l2.val;
            l2 = l2.next;
        }
        let tmp = n1 + n2 + carry;
        cur.next = new ListNode(tmp % 10);
        cur = cur.next;
        carry = Math.floor(tmp / 10); // 注意算截断的方法
    }
    if (carry > 0) {
        cur.next = new ListNode(carry);
    }
    return dummy.next;
};

# 2.两数相加(Add Two Numbers)M

# 英文题目

# 中文题目

# 模拟

2.两数相加(Add Two Numbers)M

英文题目

中文题目

模拟