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2. 两数相加

English Version

题目描述

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

解法

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        carry = 0
        dummy = ListNode(-1)
        cur = dummy
        while l1 or l2 or carry:
            s = (0 if not l1 else l1.val) + (0 if not l2 else l2.val) + carry
            carry, val = divmod(s, 10)
            cur.next = ListNode(val)
            cur = cur.next
            l1 = None if not l1 else l1.next
            l2 = None if not l2 else l2.next
        return dummy.next

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carry = 0;
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        while (l1 != null || l2 != null || carry != 0) {
            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = s / 10;
            cur.next = new ListNode(s % 10);
            cur = cur.next;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        return dummy.next;
    }
}

C++

/**
 * 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) {
        int carry = 0;
        ListNode* dummy = new ListNode(-1);
        ListNode* cur = dummy;
        while (l1 != NULL || l2 != NULL || carry != 0) {
            int s = (l1 == NULL ? 0 : l1-> val) + (l2 == NULL ? 0 : l2->val) + carry;
            carry = s / 10;
            cur->next = new ListNode(s % 10);
            cur = cur->next;
            l1 = l1 == NULL ? NULL : l1->next;
            l2 = l2 == NULL ? NULL : l2->next;
        }
        return dummy->next;
    }
};

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
  let carry = 0;
  const dummy = new ListNode(-1);
  let cur = dummy;
  while (l1 || l2 || carry) {
    const s = (l1 ? l1.val : 0) + (l2 ? l2.val : 0) + carry;
    carry = Math.floor(s / 10);
    cur.next = new ListNode(s % 10);
    cur = cur.next;
    l1 = l1 ? l1.next : l1;
    l2 = l2 ? l2.next : l2;
  }
  return dummy.next;
};

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        var carry = 0;
        while (l1 != null || l2 != null || carry != 0)
        {
            int t = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = t / 10;
            cur.next = new ListNode(t % 10);
            cur = cur.next;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        return dummy.next;
    }
}

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