Leetcode 2 Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

 

方法一：

标准链表操作，注意进位和当一个链表走完的情况。

注意最后一位如果有进位没有结算需要手动加1位。

def add_two_numbers(l1, l2)
    return l1 if l2.nil?
    return l2 if l1.nil?

    ans = ListNode.new(0)
    cur = ans
    temp = 0
    while l1 and l2
        cur.next = ListNode.new((l1.val + l2.val + temp) % 10)
        temp = (l1.val + l2.val + temp) / 10
        l1 = l1.next
        l2 = l2.next
        cur = cur.next
    end
    while l1
        cur.next = ListNode.new((l1.val + temp)%10)
        temp = (l1.val + temp) / 10
        l1 = l1.next
        cur = cur.next
    end
    while l2
        cur.next = ListNode.new((l2.val + temp)%10)
        temp = (l2.val + temp) / 10
        l2 = l2.next
        cur = cur.next
    end
    cur.next = ListNode.new(temp%10) if temp > 0
    ans.next
end

方法一优化版本：

对于一个链表走完但另一个链表没走完的情况，将该链表视为一个值为0的node处理。

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        temp = 0
        dummy, zero = ListNode(0), ListNode(0)
        cursor = dummy
        while l1 or l2:
            if not l1: l1 = zero
            if not l2: l2 = zero
            cursor.next = ListNode((l1.val + l2.val + temp) % 10)
            temp = (l1.val + l2.val + temp) / 10
            l1, l2, cursor = l1.next, l2.next, cursor.next
        if temp != 0:
            cursor.next = ListNode(temp)
        return dummy.next

 方法二：

Python、Ruby不会溢出，可以转换成数字相加后再转回链表。

def add_two_numbers(l1, l2)
    return l1 if l2.nil?
    return l2 if l1.nil?
    s1 = Array.new
    s2 = Array.new
    n = Array.new
    while l1
        s1.unshift(l1.val)
        l1 = l1.next
    end
    while l2
        s2.unshift(l2.val)
        l2 = l2.next
    end
    x = s1.join.to_i+s2.join.to_i
    n = x.to_s.chars.map(&:to_i)

    ans = ListNode.new(0)
    d = ans
    n.length.times do
        d.next = ListNode.new(n.pop)
        d = d.next
    end
    ans.next
end