###### tags: `learning` `leetcode`
# #2 Add Two numbers
https://leetcode.com/problems/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:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
## I will try: convert them to decimal numbers
- Convert l1 and l2 to numbers.
- Add them up
- Conver the sum to a linked list
---
### Try 0
```python=
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
def ltonum(l1):
l1ptr = l1
l1num = 0
while l1ptr.next != None:
l1num= l1num * 10 + l1ptr.val
l1ptr = l1ptr.next
return l1num
l1num = ltonum(l1)
l2num = ltonum(l2)
l3num = l1num + l2num
l3 = None
while l3num != 0:
l3new = ListNode(l3num % 10)
l3new.next = l3
l3num = l3num % 10
l3 = l3new
return l3
```
### try 2 wrong
```typescript=
Finished
Runtime: 8 ms
Your input
[2,4,3]
[5,6,4]
Output
[0]
Expected
[7,0,8]
```
---
### Try 1
```python=
class Solution(object):
def addTwoNumbers(self, l1, l2):
def ltonum(l1):
l1ptr = l1
l1num = 0
while l1ptr != None:
l1num= l1num * 10 + l1ptr.val
l1ptr = l1ptr.next
return l1num
l1num = ltonum(l1)
l2num = ltonum(l2)
l3num = l1num + l2num
l3 = None
while l3num != 0:
l3new = ListNode(l3num % 10)
l3new.next = l3
l3num = (l3num/ 10)
l3 = l3new
return l3
```
### try 2 wrong
```typescript=
Your input
[2,4,3]
[5,6,4]
Output
[8,0,7]
Expected
[7,0,8]
```
---
### Try 2
```python=
class Solution(object):
def addTwoNumbers(self, l1, l2):
def ltonum(l1):
l1ptr = l1
l1num = 0
while l1ptr != None:
l1num= l1num * 10 + l1ptr.val
l1ptr = l1ptr.next
return l1num
l1num = ltonum(l1)
l2num = ltonum(l2)
l3num = l1num + l2num
l3 = None
l3old = None
while l3num != 0:
l3new = ListNode(l3num % 10)
if l3old != None:
l3old.next = l3new
else:
l3 = l3new # as the head
l3num = int(l3num / 10)
l3old = l3new
return l3
```
### try 2 wrong
```typescript=
Input
[0]
[0]
Output
[]
Expected
[0]
```
#### Wrong Answer
I didn't take care of zeros.
#### Revision
I will take take of zeros
---
### Try 3
```python=
class Solution(object):
def addTwoNumbers(self, l1, l2):
def ltonum(l1):
l1ptr = l1
l1num = 0
while l1ptr != None:
l1num= l1num * 10 + l1ptr.val
l1ptr = l1ptr.next
return l1num
l1num = ltonum(l1)
l2num = ltonum(l2)
l3num = l1num + l2num
# first node
l3 = ListNode(l3num % 10) # head
# remaining nodes, if there are any
l3num = int(l3num / 10)
l3tail = l3
while l3num != 0:
l3new = ListNode(l3num % 10)
l3tail.next = l3new
l3tail = l3new
l3num = int(l3num / 10)
return l3
```
### try 3 wrong
```python=
Input
[1,8]
[0]
Output
[8,1]
Expected
[1,8]
```
#### The Mistake I made
I didn't convert l1 and l2 to numbers correctly, from the very beginning, QQQQ.
## try 4
If l1 or l2 is null, it will be mapped to a zero.
I don't need a dummy head for the output, since I will have at least a digit from the sum of l1 and l2.
```python=
class Solution(object):
def addTwoNumbers(self, l1, l2):
def ltonum(l1): # convert linked list to numbers; a null list leads to a zero
l1ptr = l1 # current position for l1
l1num = 0 # initiate the number
i = 0 # initiate the power of ten
while l1ptr != None: # to void usig a null ptr to refer
l1num += l1ptr.val * pow(10, i) # starting from least significant digit, multiplying 10^i, `i` is increasing
l1ptr = l1ptr.next # shift the current position to the next node
i += 1
return l1num
l1num = ltonum(l1)
l2num = ltonum(l2)
l3num = l1num + l2num
# first node
l3 = ListNode(l3num % 10) # head; the least significant digit as the head
# remaining nodes, if there are any
l3num = int(l3num / 10) # the next least significant digit
l3tail = l3 # only one node; the head is also the tail
while l3num != 0:
l3new = ListNode(l3num % 10)
l3tail.next = l3new # let the current tail point to the new node
l3tail = l3new # update the position of the tail
l3num = int(l3num / 10) # go to the next digit
return l3 # return the head
```
### try 4 success
Runtime: 36 ms, faster than 99.97% of Python online submissions for Add Two Numbers.
Memory Usage: 11.8 MB, less than 74.88% of Python online submissions for Add Two Numbers.
Next challenges:
Multiply Strings
Add Binary
Sum of Two Integers
Add Strings
Add Two Numbers II
Add to Array-Form of Integer
---
## learn and check
Learn to run code.
This is incorrect.
I made a mistake of reversing the order of linked list
```python=
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
def addTwoNumbers(l1, l2):
def ltonum(l1):
l1ptr = l1
l1num = 0
while l1ptr != None:
l1num= l1num * 10 + l1ptr.val
l1ptr = l1ptr.next
print(l1num)
return l1num
l1num = ltonum(l1)
l2num = ltonum(l2)
l3num = l1num + l2num
print(l3num)
l3 = None
l3old = None
while l3num != 0:
l3new = ListNode(l3num % 10)
if l3old != None:
l3old.next = l3new
else:
l3 = l3new # as the head
l3num = int(l3num / 10)
l3old = l3new
return l3
l1_2 = ListNode(2)
l1_4 = ListNode(4)
l1_3 = ListNode(3)
l1_2.next = l1_4
l1_4.next = l1_3
l2_5 = ListNode(5)
l2_6 = ListNode(6)
l2_4 = ListNode(4)
l2_5.next = l2_6
l2_6.next = l2_4
l3 = addTwoNumbers(l1_2, l2_5)
while l3 != None:
print(l3.val)
l3 = l3.next
```
```typescript=
243
564
807
7
0
8
```
## Solution of others
### Elementary Math
Java
Using a dummy head can avoid accessing a pointer with null address.
```java=
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);
ListNode p = l1, q = l2, 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 (p != null) p = p.next;
if (q != null) q = q.next;
}
if (carry > 0) {
curr.next = new ListNode(carry);
}
return dummyHead.next;
}
```
### Elementary Math, Recursive Style
Python3
```python=
class Solution:
def addTwoNumbers(self, l1, l2 ,c = 0):
val = l1.val + l2.val + c
c = val // 10
ret = ListNode(val % 10 )
if (l1.next != None or l2.next != None or c != 0):
if l1.next == None:
l1.next = ListNode(0)
if l2.next == None:
l2.next = ListNode(0)
ret.next = self.addTwoNumbers(l1.next,l2.next,c)
return ret
```
## Elementary Math, `divmod`, Dummy Head
```python=
class Solution(object):
def addTwoNumbers(self, l1, l2):
result = ListNode(0)
result_tail = result
carry = 0
while l1 or l2 or carry:
val1 = (l1.val if l1 else 0)
val2 = (l2.val if l2 else 0)
carry, out = divmod(val1+val2 + carry, 10)
result_tail.next = ListNode(out)
result_tail = result_tail.next
l1 = (l1.next if l1 else None)
l2 = (l2.next if l2 else None)
return result.next
```
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