Leetcode问题-2（Add Two Numbers）

问题描述

You are given two linked lists representing two non-negative numbers. 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.

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

分析
1. 数字非负
2. 个十百千万顺序存储
3. 返回结果链表的头结点

代码：

struct ListNode {      int val;      ListNode *next;      ListNode(int x) : val(x), next(NULL) {}};class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode *result=NULL;        ListNode *i=l1;        ListNode *j=l2;        ListNode *tail = result;        while (i != NULL || j != NULL)        {            ListNode *temp = new (ListNode)(0);            if (i != NULL && j != NULL)            {                temp->val = i->val + j->val;                i = i->next;                j = j->next;            }            else if (i != NULL && j == NULL)            {                temp->val = i->val;                i = i->next;            }            else if (i == NULL && j != NULL)            {                temp->val = j->val;                j = j->next;            }            if (result == NULL)            {                result = temp;            }            else            {                tail->next = temp;            }            tail = temp;        }        tail = result;        while (tail)        {            if (tail->val >= 10)            {                if (tail->next == NULL)                {                    tail->next = new (ListNode)(0);                }                tail->next->val += tail->val / 10;                tail->val %= 10;            }            tail = tail->next;        }        return result;    }};

代码的复杂度分析：

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空间复杂度 O(N)

• 返回了一个结果链表

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时间复杂度 O(N)

• 循环了1遍加数和被加数的链表。
• 循环了1遍结果链表。

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