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

設計

思路和平時做加法計算是一樣的。從低位往高位，兩個數的對應位與上一個低位的進位相加，得到該位的和以及進位。最高位相加後，如果還有進位，就再往高位寫下該進位。

/**
 * 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

假設l1長度為m，l2長度為n。時間複雜度就是O(max(m, n))，空間複雜度也是O(max(m, n))。 程式碼的設計還有不少不足之處。首先是SumOfTwoNumbers函式有20行，顯得有些長了；其次是buildNextNodeOfSum和addFinalCarryToSum函式使用的是輸出引數。引數多數會被當做是函式的輸入，使用輸出引數可能會造成理解上的困擾和檢查函式宣告的代價。最後，在和的連結串列構造上我的做法麻煩了一些，不如leetcode的答案精巧。