Add Two Numbers--LeetCode

最新推荐文章于 2024-08-12 10:14:14 发布

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本文介绍如何使用链表来实现两个非负整数的相加操作，包括迭代和递归两种方法，并详细解释了进位处理的技巧。

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Add Two Numbers

（原题链接：点击打开链接）

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

solution：

/**
 * 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 && l2 == NULL) return NULL;
        else if (l1 == NULL) return l2;
        else if (l2 == NULL) return l1;
        
        ListNode total(0);
        ListNode* now = &total;
        int jinwei = 0;
        ListNode* L1 = l1;
        ListNode* L2 = l2;
        
        while(L1 || L2 || jinwei)
        {
            int d1 = L1 ? L1->val : 0;
            int d2 = L2 ? L2->val : 0;
            int a = (d1 + d2 + jinwei) % 10;
            jinwei = (d1 + d2 + jinwei) / 10;
            if (L1) L1 = L1->next;
            if (L2) L2 = L2->next;
            ListNode* newnode = new ListNode(a);
            now->next = newnode;
            now = now->next;
        }
        return total.next;
    }
};

另外，我们可以用递归解决这个问题。但是在用递归的时候，要注意进位的处理，在得出的结果中要再加上一。

/**
 * 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 && l2 == NULL) return NULL;
        else if (l1 == NULL) return l2;
        else if (l2 == NULL) return l1;
        
        int num = l1->val + l2->val;
        ListNode* total = new ListNode(num % 10);
        total->next = addTwoNumbers(l1->next, l2->next);
        //处理进位
        if (num >= 10)
        {
            total->next = addTwoNumbers(total->next, new ListNode(num / 10));
        }
        return total;
    }
};