Leetcode Algorithms : 2. Add Two Numbers

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

设计

思路和平时做加法计算是一样的。从低位往高位，两个数的对应位与上一个低位的进位相加，得到该位的和以及进位。最高位相加后，如果还有进位，就再往高位写下该进位。

/**
 * 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) {
        ListNode* sum = SumOfTwoNumbers(l1, l2);
        return sum;
    }

private:
    int carry;
    bool isSoluted;

    ListNode* SumOfTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* sum = new ListNode(0);
        ListNode* top = sum;
        carry = 0;
        isSoluted = false;

        while (!isSoluted) {
            sum->val = sumOfCorrespondingBitsAndCarry(l1, l2);
            carry = sum->val / 10;
            sum->val %= 10;
            if (l1 != NULL)
                l1 = l1->next;
            if (l2 != NULL)
                l2 = l2->next;
            isSoluted = isCalculationFinished(l1, l2);
            buildNextNodeOfSum(sum);
            if (sum->next != NULL)
                sum = sum->next;
        }
        addFinalCarryToSum(sum);
        return top;
    }

    int sumOfCorrespondingBitsAndCarry(ListNode* l1, ListNode* l2) {
        int left = (l1 != NULL) ? l1->val : 0;
        int right = (l2 != NULL) ? l2->val : 0;
        return left + right + carry;
    }

    bool isCalculationFinished(ListNode* l1, ListNode* l2) {
        return (l1 == NULL && l2 == NULL);
    }

    void buildNextNodeOfSum(ListNode* sum) {
        if (!isSoluted)
            sum->next = new ListNode(0);
        else
            sum->next = NULL;
    }

    void addFinalCarryToSum(ListNode* sum) {
        if (carry != 0)
            sum->next = new ListNode(carry);
        carry = 0;
    }
};

分析

假设l1长度为m，l2长度为n。时间复杂度就是O(max(m, n))，空间复杂度也是O(max(m, n))。
代码的设计还有不少不足之处。首先是SumOfTwoNumbers函数有20行，显得有些长了；其次是buildNextNodeOfSum和addFinalCarryToSum函数使用的是输出参数。参数多数会被当做是函数的输入，使用输出参数可能会造成理解上的困扰和检查函数声明的代价。最后，在和的链表构造上我的做法麻烦了一些，不如leetcode的答案精巧。