leetcode 002
2. Add Two numbers
Difficulty:Medium
The Link
Description
You are given two non-empty linked lists representing two non-negative intergers. 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.
Solutions:
Solution A:
Brute-force attack (暴力破解)
runtime: Time limit exceeded
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l3
val1, val2 = [l1.val],[l2.val]
print (val1,val2)
while l1.next:
val1.append(l1.next.val)
while l2.next:
val2.append(l2.next.val)
# [::-1] 从后往前遍历
num1 = ''.join([str(i) for i in val1[::-1]])
num2 = ''.join([str(i) for i in val2[::-1]])
tmp = str(int(num1)+ int(num2))[::-1]
res = ListNode(int(tmp[0]))
run_res = res
for i in range(1,len(tmp)):
run_res.next = ListNode(int(tmp[i]))
run_res = run_res.next
return run_res
Solution B:
Recursion: runtime:32ms
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
if l1.val + l2.val < 10:
l3 = ListNode(l1.val+l2.val)
l3.next =self. addTwoNumbers(l1.next,l2.next)
else:
l3 = ListNode(l1.val+ l2.val - 10 )
tmp = ListNode(1)
tmp.next = None
l3.next =self. addTwoNumbers(l1.next,self.addTwoNumbers(l2.next,tmp))
return l3