Code & Func

打卡，第7天

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

从示例来看，这里的digits应该是倒过来的，即2->4->3表示的是342

如果它不是倒过来的话，我们可能还需要用栈去将元素取出来。

虽然这是一道Medium的题目，但是难度其实很小，思路大概是：

将当期指针所指向的值相加得到一个数x，那么x%10就是这个位应该为的数，x/10就是进位，所以算法思路很简单：

1
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
2

3
    int ans,add = 0;
4
    ListNode ret(0);                        //头结点让单链表操作变简单
5
    ListNode *p = &ret;
6
    while(l1 != nullptr && l2 != nullptr){
7
        ans = (l1->val + l2->val) + add;    //记得加上进位
8
        add = ans/10;                       //求出进位
9
        p->next = new ListNode(ans%10);
10
        p = p->next;
11
        l1 = l1->next;
12
        l2 = l2->next;
13
    }
14
    while(l1){
15
        ans = l1->val + add;
15 collapsed lines
16
        add = ans/10;
17
        p->next = new ListNode(ans%10);
18
        p = p->next;
19
        l1 = l1->next;
20
    }
21
    while(l2){
22
        ans = l2->val + add;
23
        add = ans/10;
24
        p->next = new ListNode(ans%10);
25
        p = p->next;
26
        l2 = l2->next;
27
    }
28
    if (add != 0) p->next = new ListNode(add);//记得出来进位不为0的情况
29
    return ret.next;
30
}

dicuss中还有一个更精炼的写法：

1
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
2
    ListNode ret(0);
3
    ListNode *p = &ret;
4
    int add = 0,sum;
5
    while(l1 || l2 || add){
6
        sum = (l1?l1->val:0) + (l2?l2->val:0) + add;
7
        add = sum/10;
8
        p->next = new ListNode(sum%10);
9
        p = p->next;
10
        l1 = (l1?l1->next:nullptr);
11
        l2 = (l2?l2->next:nullptr);
12
    }
13
    return ret.next;
14
}