LeetCode算法第二题 "add two sums"

最新推荐文章于 2021-04-15 17:31:20 发布

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最新推荐文章于 2021-04-15 17:31:20 发布

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分类专栏： LeetCode与数据结构文章标签： leetcode 算法 numbers

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本文介绍了一种算法，该算法接收两个非空链表作为输入，这两个链表分别代表两个非负整数。链表中的数字以逆序方式存储，每个节点包含一个单独的数字。文章详细阐述了如何将这两个数相加，并返回结果作为一个链表。

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Description

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode* r = (struct ListNode*)calloc(1, sizeof(struct ListNode));
    struct ListNode* p=l1;
    struct ListNode* q=l2;
    struct ListNode* s = r;
    int carry = 0;
    while(p != NULL || q != NULL){
        int x = (p != NULL) ? p->val : 0;
        int y = (q != NULL) ? q->val : 0;
        struct ListNode* t = (struct ListNode*)calloc(1, sizeof(struct ListNode));
        int sum = carry + x + y;
        carry = sum / 10;
        t->val = sum % 10;
        t->next = NULL;
        s->next = t;
        s = t;
        if (p != NULL)
            p = p->next;
        if (q != NULL)
            q = q->next;
    }
    if(carry == 1){
        struct ListNode* t = (struct ListNode*)calloc(1, sizeof(struct ListNode));
        t->val = 1;
        t->next = NULL;
        s->next = t;
    } 
    return r->next;
}

Conclusion

How to create the target list without knowing the actual lengths of the input lists?
Be careful when dealing with the carry.
How to return the correct pointer of the target list? By adding a head of the target list and return from its next.