This data structure is about linked list.
The origianl question is as follow.
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.
This can be solved by setting the flag to count if a number is carry or not.
We should judge the length of the two list. A good method is looping the two list together. Which stop faster means the list is shorter.
This is my solution.
# 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
"""
flag = 0
if l1 == None: return l2
if l2 == None: return l1
dummy = ListNode(0); res = dummy
while l1 and l2:
res.next = ListNode((l1.val+l2.val+flag) % 10)
flag = (l1.val+l2.val+flag) / 10
l1 = l1.next
l2 = l2.next
res = res.next
if l2:
while l2:
res.next = ListNode((l2.val+flag) % 10)
flag = (l2.val+flag) / 10
l2 = l2.next
res = res.next
if l1:
while l1:
res.next = ListNode((l1.val+flag) % 10)
flag = (l1.val+flag) / 10
l1 = l1.next
res = res.next
if flag == 1: res.next = ListNode(1)
return dummy.next
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