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