LeetCode 002 AddTwoNumbers

于 2016-12-12 18:57:59 发布
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分类专栏： LeetCode 文章标签： C++ Leetcode
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
本文链接：https://blog.youkuaiyun.com/lizi_stdio/article/details/53585922
本文介绍如何解决两个以链表形式表示的非负整数相加的问题，并返回结果作为一个新的链表。通过分析两种不同的实现方法，一种是不使用头节点直接进行操作，另一种则是引入头节点简化插入过程。

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2. Add Two Numbers

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
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

    }
};


解题思路：
  • 自己的解题思路
首先，依次遍历l1跟l2，然后再遍历其中较长的链表，最后注意进位的处理。逻辑比较简单，但是每次都要分开讨论是不是第一次插入。
  • 别人的解题思路
先建立一个头结点，带来的好处：统一插入过程。
学习收获：
  • 第一次尝试处理链表类型的题目；知道，头结点的好处。
  • 对于进位，常用的变量名是carry。
附件1：程序
1、自己的程序：
这个程序会出现TLE错误，所以需要修改一下。由于没有考虑分开插入，所以出现红色标记的错误。
/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
    public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
    {
        int temp = 0;
        ListNode* l3 = NULL;
        ListNode* p = l3;
        while(l1&&l2)
        {
            int sum = l1->val + l2->val + temp;
            temp = sum / 10;
            ListNode* l3_ptr = new ListNode(sum % 10);
            if(l3 == NULL)
            {
                l3 = l3_ptr;
                p = l3;
            }
            p->next = l3_ptr;
            p = p->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1)
        {
            int sum = l1->val + temp;
            temp = sum / 10;
            ListNode* l3_ptr = new ListNode(sum % 10);
            if(l3 == NULL)
            {
                l3 = l3_ptr;
                p = l3;
            }
            p->next = l3_ptr;
            p = p->next;
            l1 = l1->next;
        }
        while(l2)
        {
            int sum = l2->val + temp;
            temp = sum / 10;
            ListNode* l3_ptr = new ListNode(sum % 10);
            if(l3 == NULL)
            {
                l3 = l3_ptr;
                p = l3;
            }
            p->next = l3_ptr;
            p = p->next;
            l2 = l2->next;
        }
        if(temp)
        {
            ListNode* l3_ptr = new ListNode(temp);
            if(l3 == NULL)
            {
                l3 = l3_ptr;
                p = l3;
            }
            p->next = l3_ptr;
        }
        return l3;
    }
};
改正后的程序：
/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
    public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
    {
        int temp = 0;
        ListNode* l3 = NULL;
        ListNode* p = l3;
        while(l1&&l2)
        {
            int sum = l1->val + l2->val + temp;
            temp = sum / 10;
            ListNode* l3_ptr = new ListNode(sum % 10);
            if(l3 == NULL)
            {
                l3 = l3_ptr;
                p = l3;
            }
            else
            {
                p->next = l3_ptr;
                p = p->next;
            }
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1)
        {
            int sum = l1->val + temp;
            temp = sum / 10;
            ListNode* l3_ptr = new ListNode(sum % 10);
            if(l3 == NULL)
            {
                l3 = l3_ptr;
                p = l3;
            }
            else
            {
                p->next = l3_ptr;
                p = p->next;
            }
            l1 = l1->next;
        }
        while(l2)
        {
            int sum = l2->val + temp;
            temp = sum / 10;
            ListNode* l3_ptr = new ListNode(sum % 10);
            if(l3 == NULL)
            {
                l3 = l3_ptr;
                p = l3;
            }
            else
            {
                p->next = l3_ptr;
                p = p->next;
            }
            l2 = l2->next;
        }
        if(temp)
        {
            ListNode* l3_ptr = new ListNode(temp);
            if(l3 == NULL)
            {
                l3 = l3_ptr;
                p = l3;
            }
            p->next = l3_ptr;
        }
        return l3;
    }
};
在看这题的discuss区的时候，发现可以写一个头指针，所以又把自己的程序改了一下。
/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
    public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int carry = 0;
        ListNode *list = new ListNode(0);
        ListNode *head = list;
        ListNode *prev = list;
        while(l1 && l2)
        {
            carry += l1->val + l2->val;
            list = new ListNode(0);
            list->val = carry % 10;
            carry /= 10;
            l1 = l1->next;
            l2 = l2->next;
            prev->next = list;
            prev = prev->next;
        }
        while(l1)
        {
            carry += l1->val;
            list = new ListNode(0);
            list->val = carry % 10;
            carry /= 10;
            l1 = l1->next;
            prev->next = list;
            prev = prev->next;
        }
        while(l2)
        {
            carry += l2->val;
            list = new ListNode(0);
            list->val = carry % 10;
            carry /= 10;
            l2 = l2->next;
            prev->next = list;
            prev = prev->next;
        }
        if(carry)
        {
            list = new ListNode(0);
            list->val = carry;
            prev->next = list;
            prev = prev->next;
        }
        return head->next;
    }
};
2、别人的程序
不带头结点的：
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
{
    ListNode *head = NULL, *prev = NULL;
    int carry = 0;
    while(l1 || l2)
    {
        int v1 = l1? l1->val: 0;
        int v2 = l2? l2->val: 0;
        int tmp = v1 + v2 + carry;
        carry = tmp / 10;
        int val = tmp % 10;
        ListNode* cur = new ListNode(val);
        if(!head) head = cur;
        if(prev) prev->next = cur;
        prev = cur;
        l1 = l1? l1->next: NULL;
        l2 = l2? l2->next: NULL;
    }
    if(carry > 0)
    {
        ListNode* l = new ListNode(carry);
        prev->next = l;
    }
    return head;
}
带头结点的：
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
{
    ListNode preHead(0), *p = &preHead;
    int extra = 0;
    while(l1 || l2 || extra)
    {
        if(l1) extra += l1->val, l1 = l1->next;
        if(l2) extra += l2->val, l2 = l2->next;
        p->next = new ListNode(extra % 10);
        extra /= 10;
        p = p->next;
    }
    return preHead.next;
}
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