LeetCode 之 Add Two Numbers — C 实现

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

给定两个存有非负正数的链表，整数的每位数字按序存储在每个节点中，将两个数相加并以链表形式返回。

分析：

    注意两个链表不一样长的情况，以及进位，最高位有进位的需要新分配一个节点。

<pre name="code" class="html">struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {    struct ListNode *head =NULL, *pNode = NULL, *cur = NULL;    int sum=0, carry = 0;        if((NULL == l1)||(NULL == l2))        return NULL;    while((NULL != l1)&&(NULL != l2))    {        pNode= (struct ListNode *)malloc(sizeof(struct ListNode));        sum = l1->val + l2->val + carry;        carry = sum/10;        pNode->val = sum%10;        pNode->next = NULL;        if(NULL == head)        {            head = cur = pNode;        }        else        {            cur->next = pNode;            cur = pNode;        }        l1 = l1->next;        l2 = l2->next;    }    if(NULL != l1)    {        while(NULL != l1)        {            pNode= (struct ListNode *)malloc(sizeof(struct ListNode));            sum = l1->val + carry;            carry = sum/10;            pNode->val = sum%10;            pNode->next = NULL;                       cur->next = pNode;            cur = pNode;            l1 = l1->next;        }    }    if(NULL != l2)    {        while(NULL != l2)        {            pNode= (struct ListNode *)malloc(sizeof(struct ListNode));            sum = l2->val + carry;            carry = sum/10;            pNode->val = sum%10;            pNode->next = NULL;                       cur->next = pNode;            cur = pNode;            l2 = l2->next;        }    }    if(carry != 0)    {        pNode= (struct ListNode *)malloc(sizeof(struct ListNode));        pNode->val = carry;        pNode->next = NULL;               cur->next = pNode;    }        return head;}