README_EN.md

02.05. Sum Lists

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Description

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

 

Example:

Input: (7 -> 1 -> 6) + (5 -> 9 -> 2). That is, 617 + 295.

Output: 2 -> 1 -> 9. That is, 912.

Follow Up: Suppose the digits are stored in forward order. Repeat the above problem.

Example:

Input: (6 -> 1 -> 7) + (2 -> 9 -> 5). That is, 617 + 295.

Output: 9 -> 1 -> 2. That is, 912.

Solutions

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;
};

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