問題描述

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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

【解法一】

模擬十進制加法，兩兩合并饺律，當一個list為空時復制領一個list即可窃页，此時需注意處理最后進位的1

/**
 * 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 listSum(0);
        ListNode *pSum = &listSum;
        int flag = 0;
        int sum = 0;
        while ((l1!=NULL) && (l2!=NULL))
        {
            sum = l1->val+ l2->val + flag;
            // 計算是否要進位
            flag = sum/10;
            // 計算當前單位的值
            pSum->next = new ListNode(sum%10);
            // 指針跳轉下一個節(jié)點
            pSum = pSum->next;
            // l1和l2推進
            l1 = l1->next;
            l2 = l2->next;
        }
        while (l1!=NULL)
        {
            sum = l1->val + flag;
            flag = sum/10;
            pSum->next = new ListNode(sum%10);
            pSum = pSum->next;
            l1 = l1->next;
        }
        while (l2!=NULL)
        {
            sum = l2->val + flag;
            flag = sum/10;
            pSum->next = new ListNode(sum%10);
            pSum = pSum->next;
            l2 = l2->next;
        }
        if (flag)
        {
            pSum->next = new ListNode(flag);
        }
        return listSum.next;
    }
};

【解法二】

由解法可以看出來，每一步的結果實際上是相似的复濒，只是省略為NULL和為0的情況脖卖，所以代碼可以簡化為如下形式：

/**
 * 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 listSum(0);
        ListNode *pSum = &listSum;
        int flag = 0;
        int sum = 0;
        while ((l1!=NULL) || (l2!=NULL) || flag)
        {
            sum = (l1?l1->val:0) + (l2?l2->val:0) + flag;
            // 計算是否要進位
            flag = sum/10;
            // 計算當前單位的值
            pSum->next = new ListNode(sum%10);
            // 指針跳轉下一個節(jié)點
            pSum = pSum->next;
            // l1和l2推進
            l1 = l1 ?l1->next:l1;
            l2 = l2 ?l2->next:l2;
        }
        return listSum.next;
    }
};