2. Add Two Numbers (Medium)

Q:

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 contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

 

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

 

Constraints:

• The number of nodes in each linked list is in the range [1, 100].
• 0 <= Node.val <= 9
• It is guaranteed that the list represents a number that does not have leading zeros.

A:

Mistakes:
1: 忘记 给l1, l2 也增加runner，    ---------   这里其实可以直接用l1,l2当做runner，毕竟java是传引用的。

2： 应该是在计算完毕 sun之后，再update carry的值。
3: 当carry 最后 进一位的时候， 我们copy 其余的产品的时候，  忘了把其也加上。 哎~~~~~
  这道题，就是考的细心啊~~~~~~~~~
4: 为什么，（ 3+5 )%10 的结果总是1 呢?  搞得carry 变成1 了。 哎 ---------- 计算carry的时候，把r2 写成r1了。

5: 当carry 最后加完了， 还是1 的时候，我们要再生成一个节点。   ╮(╯▽╰)╭，  Tiger，你这样不行啊~~~~~~~~~~~~SB错误太多了

--------------------------3 rd pass--------------use only one header,   for newly created list header 跟第一遍的写法类似----------------------  好消息是，  只有几个拼写错误，  不好的是， 还是不能bug free--------------------

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int carry = 0;
        ListNode header(0);
        auto tail = &header;
        while (carry || l1 || l2) {
            int a = l1 ? l1->val : 0;
            int b = l2 ? l2->val : 0;
            tail->next = new ListNode((a + b + carry) % 10);
            tail = tail->next;
            if (l1) {
                l1 = l1->next;
            }
            if (l2) {
                l2 = l2->next;
            }
            carry = (a + b + carry) / 10;
        }
        return header.next;
    }
};

-------------------4 th pass---------------recursive way---------------

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return helper(l1,l2,0);
    }
    private ListNode helper(ListNode l1,ListNode l2, int carry){
        ListNode curNode;
        if(l1==null && l2 == null){
            return carry == 0?null: new ListNode(carry);
        }else if(l1 == null){
            curNode = new ListNode((carry+l2.val)%10);
            carry = (carry+l2.val)/10;
            curNode.next = helper(null,l2.next,carry);
        }else if(l2 == null){
            curNode = new ListNode((carry+l1.val)%10);
            carry = (carry+l1.val)/10;
            curNode.next = helper(l1.next,null,carry);
        }else{
            curNode = new ListNode((carry+l1.val+l2.val)%10);
            carry = (carry+l1.val+l2.val)/10;
            curNode.next = helper(l1.next,l2.next,carry);
        }
        return curNode;
    }
}

----------------------递归调用的  简化版本------------

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return helper(l1,l2,0);
    }
    private ListNode helper(ListNode l1,ListNode l2, int carry){     
        if(l1==null && l2 == null)
            return carry == 0?null: new ListNode(carry);        
        int sum = (l1==null?0:l1.val) + (l2==null?0:l2.val)+carry;
        ListNode curNode = new ListNode((sum)%10);
        curNode.next = helper(l1==null?null:l1.next,l2==null?null:l2.next,sum/10);
        return curNode;
    }
}