Two-sum

two-sum
题目描述
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
解题思路
思路非常简单，利用两个指针分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

/** * 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 carry=0;        ListNode* tail=new ListNode(0);        ListNode* ptr=tail;        while(l1 !=NULL|| l2 != NULL)        {            int val1=0;            if(l1 != NULL)            {                val1 = l1->val;                l1=l1->next;            }            int val2 = 0;            if(l2 != NULL)            {                val2 = l2->val;                l2 = l2->next;            }        int tmp = val1 + val2 + carry;        ptr->next = new ListNode(tmp%10);        carry = tmp/10;        ptr = ptr->next;    }    if(carry == 1)    {        ptr->next = new ListNode(1);    }    return tail->next;    }};

在链表的编程中，常常需要新建一个头，然后用一个该类型链表的指针指向该头，在进行随后的各种操作时，对该指针进行操作即可。如果不这么做，会导致链表信息的缺失，找不到链表的头。

/** * 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 carry=0;        ListNode* tail=new ListNode(0);        //ListNode* ptr=tail;        while(l1 !=NULL|| l2 != NULL)        {            int val1=0;            if(l1 != NULL)            {                val1 = l1->val;                l1=l1->next;            }            int val2 = 0;            if(l2 != NULL)            {                val2 = l2->val;                l2 = l2->next;            }        int tmp = val1 + val2 + carry;        tail->next = new ListNode(tmp%10);        carry = tmp/10;        tail = tail->next;    }    if(carry == 1)    {        tail->next = new ListNode(1);    }    return tail;    }};

此时，没有使用ptr指针，在不断的tail = tail->next之后，tail就将指向最后一个节点，此时，return tail 就只返回最后一个节点。