[leetcode-2-Add Two Numbers]

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.

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

刚开始ac的代码如下：简直是啰嗦到家

就是两个链表从头到尾非空的情况下往前计算

提防着进位等等 还要处理最后一个为空的边界条件

运行效率简直惨不忍睹：先贴这儿。

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
    {
        int carry = 0;//进位
        ListNode* result = l1;
        while (l1->next != NULL && l2->next != NULL)
        {
            l1->val = l1->val + l2->val + carry;                            
            if (l1->val >= 10)
            {
                l1->val -= 10;
                carry = 1;
            }
            else carry = 0;

            l1 = l1->next;
            l2 = l2->next;
        }
        l1->val = l1->val + l2->val + carry;
        if (l1->val >= 10)
        {
            l1->val -= 10;
            carry = 1;
        }
        else carry = 0;
        if (l1->next == NULL && l2->next == NULL)
        {
            if (carry == 1)
            {
                ListNode* newNode = new ListNode(1);
                l1->next = newNode;
            }
        }
        else
        {            
            if (l1->next == NULL &&  l2->next != NULL)//l2非空
            {            
                l1->next = l2->next;
                l1 = l1->next;
                while (carry && l1->next!=NULL)
                {
                    l1->val += carry;
                    if (l1->val >= 10)
                    {
                        l1->val -= 10;
                        carry = 1;
                    }
                    else carry = 0;
                    l1 = l1->next;
                }
                if (carry ==1)//最后一位
                {
                    l1->val += carry;
                    if (l1->val >= 10)
                    {
                        l1->val -= 10;
                        ListNode* newNode = new ListNode(1);//新建结点
                        l1->next = newNode;
                        carry = 1;
                    }
                    else carry = 0;
                }
                
            }
            else if (l1->next != NULL && l2->next == NULL)
            {                
                l1 = l1->next;
                while (carry && l1->next != NULL)
                {
                    l1->val += carry;
                    if (l1->val >= 10)
                    {
                        l1->val -= 10;
                        carry = 1;
                    }
                    else carry = 0;
                    l1 = l1->next;
                }
                if (carry == 1)//最后一位
                {
                    l1->val += carry;
                    if (l1->val >= 10)
                    {
                        l1->val -= 10;
                        ListNode* newNode = new ListNode(1);//新建结点
                        l1->next = newNode;
                        carry = 1;
                    }
                    else carry = 0;
                }
            }
        }    
        
        return result;
    }

然后参考别人大牛的代码

简洁到家啊~~~ 如下：

 public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode c1 = l1;
        ListNode c2 = l2;
        ListNode sentinel = new ListNode(0);
        ListNode d = sentinel;
        int sum = 0;
        while (c1 != null || c2 != null) {
            sum /= 10;
            if (c1 != null) {
                sum += c1.val;
                c1 = c1.next;
            }
            if (c2 != null) {
                sum += c2.val;
                c2 = c2.next;
            }
            d.next = new ListNode(sum % 10);
            d = d.next;
        }
        if (sum / 10 == 1)
            d.next = new ListNode(1);
        return sentinel.next;
    }

不光简洁 效率也还过得去