[LeetCode] 2. Add Two Numbers 两个数字相加

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.

给定两个链表分别代表两个非负整数。数位以倒序存储，并且每一个节点包含一位数字。将两个数字相加并以链表形式返回。题目不难，主要考察链表操作和加法进位。

进位 = sum / 10，当前位值 = sum % 10

Java:

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        int carry = 0;
        while (l1 != null || l2 != null) {
            int d1 = l1 == null ? 0 : l1.val;
            int d2 = l2 == null ? 0 : l2.val;
            int sum = d1 + d2 + carry;
            carry = sum >= 10 ? 1 : 0;
            cur.next = new ListNode(sum % 10);
            cur = cur.next;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }
        if (carry == 1) cur.next = new ListNode(1);
        return dummy.next;
    }
}

Python: wo

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        carry = 0
        dummy = ListNode(0)
        temp = dummy
        while l1 or l2:
            vals = carry
            if l1:
                vals += l1.val
                l1 = l1.next
            if l2:
                vals += l2.val
                l2 = l2.next
            temp.next = ListNode(vals % 10)
            temp = temp.next
            carry = vals / 10
            
        if carry:
            temp.next = ListNode(carry)
            
        return dummy.next    　　

Python:

class ListNode:
    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
        """
        dummy = ListNode(0)
        current, carry = dummy, 0

        while l1 or l2:
            val = carry
            if l1:
                val += l1.val
                l1 = l1.next
            if l2:
                val += l2.val
                l2 = l2.next
            carry, val = divmod(val, 10)
            current.next = ListNode(val)
            current = current.next

        if carry == 1:
            current.next = ListNode(1)

        return dummy.next

if __name__ == '__main__':
    a, a.next, a.next.next = ListNode(2), ListNode(4), ListNode(3)
    b, b.next, b.next.next = ListNode(5), ListNode(6), ListNode(4)
    result = Solution().addTwoNumbers(a, b)
    print "{0} -> {1} -> {2}".format(result.val, result.next.val, result.next.next.val)

Python:

class Solution:
    # @return a ListNode
    def addTwoNumbers(self, l1, l2):
        head = ListNode(0)
        l = head
        carry = 0
        while l1 or l2 or carry:
            sum, carry = carry, 0
            if l1:
                sum += l1.val
                l1 = l1.next
            if l2:
                sum += l2.val
                l2 = l2.next
            if sum > 9:
                carry = 1
                sum -= 10
            l.next = ListNode(sum)
            l = l.next
        return head.next

C++:

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode *res = new ListNode(-1);
        ListNode *cur = res;
        int carry = 0;
        while (l1 || l2) {
            int n1 = l1 ? l1->val : 0;
            int n2 = l2 ? l2->val : 0;
            int sum = n1 + n2 + carry;
            carry = sum / 10;
            cur->next = new ListNode(sum % 10);
            cur = cur->next;
            if (l1) l1 = l1->next;
            if (l2) l2 = l2->next;
        }
        if (carry) cur->next = new ListNode(1);
        return res->next;
    }
};