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;    }

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