leetcode 2 Add Two Numbers（链表）

题目：

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.

思路：

题目主要考察链表，利用链表来模拟实现加法，容易出错误的地方在于对加法进位的处理。我自己的思路比较复杂，主要是考虑当l1,l2其中有一个为null时，将进位1转移到l1或者l2，转移后如果仍有一个为Null，则可直接退出循环。

 

代码：

struct ListNode {
  int val;
  ListNode *next;
  ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int left = 0;
        ListNode* first = new struct ListNode(0);
        ListNode* now = first;
        while(l1 != NULL && l2 != NULL){
            int temp = l1->val + l2->val + left;
            now->val = temp % 10;
            if(temp >= 10)
                left = 1;
            else
                left = 0;
            l1 = l1->next;
            l2 = l2->next;
            if(l1 == NULL && left == 1){
                ListNode* newNode = new struct ListNode(1);
                l1 = newNode;
                left = 0;
            }
            else if(l2 == NULL && left == 1){
                ListNode* newNode = new struct ListNode(1);
                l2 = newNode;
                left = 0;
            }
            if(l1 != NULL && l2 != NULL)
            {
                ListNode* nextNode = new struct ListNode(0);
                now->next = nextNode;
                now = now->next;
            }
        }
        if(l1 == NULL)
            now->next = l2;
        else
            now->next = l1;
        return first;
    }
};

网上的一种思路是将三者同等看待，只要l1,l2其中有一个不为null，就可以接着计算。

代码比较简洁，也贴一下：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

复杂度分析：

讲道理，我的复杂度是min(l1,l2)，网上思路的复杂度是max(l1,l2)。不过都是线性的，基本也没啥区别吧。