2. Add Two Numbers

Question

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.

Solution

• 基于链表的运算
• 逐位加法：计算当前位，更新进位，更新操作数
• 当前位有至少一个操作数 or 有进位的时候，结果的当前位需要进行计算
  • 即l1!=NULL||l2!=NULL||carry==1
• 链表需要有一个头部指针（result）和一个滑动指针（curr）
• 更为简洁地写法，使用三目运算符和%操作
  • C++可以直接将指针作为逻辑表达式，NULL返回false，非NULL返回true

更为简洁地写法：

/**
 * 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 = new ListNode(0);
        ListNode *curr = sum;
        int carry = 0;
        while (l1!=NULL || l2!=NULL || carry!=0) {
            curr->next = new ListNode(0);
            curr = curr->next;
            
            curr->val = (l1?l1->val:0)+(l2?l2->val:0)+carry;
            carry = curr->val>=10?1:0;
            curr->val = curr->val%10;
            
            l1 = l1?l1->next:l1;
            l2 = l2?l2->next:l2;
        }
        return sum->next;
    }
};

/**
 * 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 carry = 0;
        ListNode* result = new ListNode(0);
        ListNode* curr = fresult;
        for(;l1!=NULL||l2!=NULL||carry==1;) {
          	// 如果满足条件，为结果新创建一位
            curr->next = new ListNode(0);
            curr = curr->next;
            curr->val += carry;
            if (l1!=NULL) {
                curr->val += l1->val;
                l1 = l1->next;
            }
            if (l2!=NULL) {
                curr->val += l2->val;
                l2 = l2->next;
            }
            
            if (curr->val >= 10) {
                carry = 1;
                curr->val -= 10;
            }
            else {
                carry = 0;
            }
        }
        return result->next;
    }
};