# 2.Add-Two-Numbers

## 2. Add Two Numbers

## 题目地址

<https://leetcode.com/problems/add-two-numbers/>

## 题目描述

```
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:

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

## 代码

### Approach #1

Elementary Math Intuition Keep track of the carry using a variable and simulate digits-by-digits sum starting from the head of list, which contains the least-significant digit.

Complexity Analysis

* Time complexity : O(max(m, n)). Assume that m and n represents the length of l1 and l2 respectively, the algorithm above iterates at most max(*m*,*n*) times.
* Space complexity : &#x4F;*(max(\_m*,*n*)). The length of the new list is at most max(*m*,\_n)+1.

```java
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2;
    ListNode curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
      int x = (p != null) ? p.val : 0;
      int y = (q != null) ? q.val : 0;
      int sum = carry + x + y;
      carry = sum / 10;
      curr.next = new ListNode(sum % 10);
      curr = curr.next;
      if (q != null) p = p.next;
      if (q != null) q = q.next;
    }

    if (carry > 0) {
      curr.next = new ListNode(carry);
    }

    return dummyHead.next;
  }
}
```


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