笔试题15. LeetCode OJ (2)

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2. Add Two Number

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) {        if(l1 == NULL && l1 == NULL)        {             return NULL;         }        if(l1 == NULL)        {             return l2;         }        if(l2 == NULL)        {              return l1;         }                bool ishead = true;        int increase=0;        ListNode *newHead = NULL;        ListNode *cur = NULL;        while(l1 && l2)        {            ListNode * tmp = new ListNode(l1->val+l2->val);            if(increase != 0)            {                tmp->val+=increase;            }                        increase = tmp->val / 10;            tmp->val %= 10;                        if(cur != NULL)            {                cur->next = tmp;                cur = cur->next;            }                        if(ishead)            {                newHead = tmp;                cur = newHead;                ishead = false;            }                        l1=l1->next;            l2=l2->next;        }                while(l1)        {            cur->next = new ListNode(l1->val);            cur = cur->next;            if(increase != 0)            {                cur->val+=increase;                increase = cur->val/10;                cur->val%=10;            }            l1=l1->next;        }        while(l2)        {            cur->next = new ListNode(l2->val);            cur = cur->next;            if(increase != 0)            {                cur->val+=increase;                increase = cur->val/10;                cur->val%=10;            }            l2=l2->next;           }                if(increase != 0)        {            cur->next = new ListNode(increase);        }                return newHead;    }};

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