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.

Example

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Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8
  
Explanation: 342 + 465 = 807.

Interface

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/**
 * 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) {
        
    }
};

Solution

题目挺简单的，就是通过链表实现大数加法，结果的每一位由三个值决定，两个是两个加数在这个数位上的值，剩下一个是上一位的进位，主要就是需要注意进位的实现，也没什么难度，可能还有需要注意的就是代码的逻辑需要简明清晰，不然很容易写的很冗长。

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/**
 * 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) {
        int n1 = 0, n2 = 0, tmp = 0, carry = 0;
        ListNode *p1 = l1, *p2 = l2, *temp = NULL, *result = NULL;
        while (p1 != NULL || p2 != NULL || carry != 0) {
            n1 = (p1 != NULL) ? p1->val : 0;
            n2 = (p2 != NULL) ? p2->val : 0;
            tmp = n1 + n2 + carry;
            carry = tmp / 10;
            if (result == NULL) result = new ListNode(tmp % 10);
            else temp->next = new ListNode(tmp % 10);
            p1 = (p1 == NULL) ? p1 : p1->next;
            p2 = (p2 == NULL) ? p2 : p2->next;
            if (temp == NULL) temp = result;
            else temp = temp->next;
        }
        return result;
    }
};

Improvement

在 LeetCode 讨论区中发现了更简洁的代码。

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
ListNode* addTwoNumbers(ListNode *l1, ListNode *l2) {
    ListNode preHead(0), *p = &preHead;
    int extra = 0;
    while (l1 || l2 || extra) {
        int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;
        extra = sum / 10;
        p->next = new ListNode(sum % 10);
        p = p->next;
        l1 = l1 ? l1->next : l1;
        l2 = l2 ? l2->next : l2;
    }
    return preHead.next;
}