Add Two Numbers

问题描述

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.

示例

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

算法思路

在这个问题中，有两种解决思路，第一种是先将输入的链表计算结果转换成字符串，再转成链表输出；第二种是直接采用链表操作。总的来说，第一种方法更易于理解，执行效率更高（runtime is 116ms），第二种方法更加复杂，耗时更长(runtime is 288ms),所以推荐使用方法一。

代码实现（Python3）

方法一：

class Solution:
    def addTwoNumbers(self, l1, l2):
        twoSum = str(int(self.getNumber(l1)) + int(self.getNumber(l2)))
        return self.constructList(twoSum)
    
    def getNumber(self, head):
        current, res = head, []
        while current:
            res.append(str(current.val))
            current = current.next
        return ''.join(res[::-1]) # reverse res

    def constructList(self, num):
        head = ListNode(int(num[-1])) # get the last element
        current = head
        for i in range(len(num) - 2, -1, -1): # range(start, stop, step)
            node = ListNode(int(num[i]))
            current.next = node
            current = current.next
        return head

方法二：

# 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
        """
        
        # Carry over place holder.
        carry = 0
        
        # Create a dummy head, tail node to keep track of the start and end of the list
        dummy_head = tail = ListNode(0)
        
        # While there are elements in either list we add them
        while l1 or l2:
            
            # Get the two element values, if there is not a node use 0
            num_one = l1.val if l1 else 0
            num_two = l2.val if l2 else 0
            
            # Get the new total with two numbers plus the carry over
            new_sum = num_one + num_two + carry
            
            # Check to see if the number will create a carry
            if new_sum > 9:
                # Set the tail node equal to the new node.
                tail.next = ListNode(new_sum - 10)
                # Set a carry because the number was larger than 10
                carry = 1
            else:
                tail.next = ListNode(new_sum)
                carry = 0
                
            # Set the tail node to the new tail.
            tail = tail.next 
            
            # Set the current nodes to the next number
            if l1:
                l1 = l1.next
            if l2:
                l2 = l2.next
        
        # Once the adding has been completed if there is still a carry append a new node
        if carry:
            tail.next = ListNode(carry)
            
        # Return the first number set, dummy head was just a place holder. The tail would point to the last node and not the first number in the linked list.
        return dummy_head.next

参考资料

1. https://stackoverflow.com/questions/31633635/what-is-the-meaning-of-inta-1-in-python What is the meaning of “int(a[::-1])” in Python?
2. https://stackoverflow.com/questions/930397/getting-the-last-element-of-a-list-in-python Getting the last element of a list in Python
3. https://www.pythoncentral.io/pythons-range-function-explained/ Python’s range() Function Explained
4. https://leetcode.com/problems/add-two-numbers/ Add Two Numbers