2-Add Two Numbers @LeetCode

题目

思路

题目中得到的信息有：

1. 这是两个非负数，每位分别保存在链表的一个结点上；
2. 逆序保存，从低位到高位依次。

一般整数的相加都是从低往高进行，和保存的顺序一致，因此一次遍历就可完成，可以看出这道题目不难。

C算法

/** * Definition for singly-linked list.  * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {    struct ListNode *ret, *p, *prev;    p = ret = prev = NULL;    int flag = 0;    //先是两个整数相加    while((l1 != NULL) && (l2 != NULL) ) {         p = (struct ListNode *)malloc(sizeof(struct ListNode));        p->val = l1->val + l2->val + flag;        flag = p->val / 10;        p->val -= (flag > 0 ? 10 : 0);        if (prev != NULL) prev->next = p;        prev = p;        if (ret == NULL) ret = p;        l1 = l1->next;        l2 = l2->next;    }    //若是l1还有结点，添加上去    while(l1 != NULL) {        p = (struct ListNode *)malloc(sizeof(struct ListNode));         p->val = l1->val  + flag;         flag = p->val / 10;        p->val -= (flag > 0 ? 10 : 0);        l1 = l1->next;          if (prev != NULL) prev->next = p;        if (ret == NULL) ret = p;        prev->next = p;        prev = p;    }    //若是l2还有结点，添加上去    while(l2 != NULL) {        p = (struct ListNode *)malloc(sizeof(struct ListNode));         p->val = l2->val  + flag;         flag = p->val / 10;        p->val -= (flag > 0 ? 10 : 0);        l2 = l2->next;          if (prev != NULL) prev->next = p;         if (ret == NULL) ret = p;        prev->next = p;        prev = p;    }    //最后可能会有进位，要考虑到    if (flag != 0) {        p = (struct ListNode *)malloc(sizeof(struct ListNode));        p->val = flag;        prev->next = p;        prev = p;        flag = 0;    }    prev->next = NULL;    return ret;}

结果