LeetCode: 2. Add Two Numbers

题目描述

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:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

解题思路 —— 递归求解

一般遇到链表和树相关的问题，我都习惯用递归求解。因为他们的定义本身就是递归的，非常适合递归求解。
这道题只需要将按顺序遍历两条节点，相加构造和的节点，同时将进位用于构造和的下一个节点。

AC 代码

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */
class Solution {
private:
    // 求两个链表的和，其中 carryNum 是前面的进位
    ListNode* addTwoNumbersWithCarry(ListNode* l1, ListNode* l2, int carryNum) {
        if(l1 == nullptr && l2 == nullptr && carryNum == 0){
            return nullptr;
        }

        // 计算当前节点
        ListNode* pNumber = new ListNode(carryNum);
        if(l1 != nullptr) {
            pNumber->val += l1->val;
            l1 = l1->next;
        }
        if(l2 != nullptr){
            pNumber->val += l2->val;
            l2 = l2->next;
        }

        // 计算后面的节点
        pNumber->next = addTwoNumbersWithCarry(l1, l2, pNumber->val/10);
        pNumber->val %= 10;

        return pNumber;
    }
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        return addTwoNumbersWithCarry(l1, l2, 0);
    }
};