2. Add Two Numbers

题目

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.

题目大意

2 个逆序的链表，要求从低位开始相加，得出结果也逆序输出，返回值是逆序结果链表的头结点。

解题思路

需要注意的是各种进位问题。

极端情况，例如

Input: (9 -> 9 -> 9 -> 9 -> 9) + (1 -> )
Output: 0 -> 0 -> 0 -> 0 -> 0 -> 1

为了处理方法统一，可以先建立一个虚拟头结点，这个虚拟头结点的 Next 指向真正的 head，这样 head 不需要单独处理，直接 while 循环即可。另外判断循环终止的条件不用是 p->next ！= null，这样最后一位还需要额外计算，循环终止条件应该是 p != nil。

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
 public:
  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode dummy_head;
    ListNode* p = &dummy_head;
    ListNode* p1 = l1;
    ListNode* p2 = l2;
    int sum;
    int carry = 0;
    while (p1 != nullptr && p2 != nullptr) {
      sum = p1->val + p2->val + carry;      
      carry = sum / 10;
      p1 = p1->next;
      p2 = p2->next;
      p->next = new ListNode(sum % 10);
      p = p->next;
    }
    while (p1 != nullptr) {
      sum = p1->val + carry;
      p1 = p1->next;
      carry = sum / 10;
      p->next = new ListNode(sum % 10);
      p = p->next;
    }
    while (p2 != nullptr) {
      sum = p2->val + carry;
      p2 = p2->next;
      carry = sum / 10;
      p->next = new ListNode(sum % 10);
      p = p->next;
    }
    if (carry) {
      p->next = new ListNode(carry);
      p = p->next;
    }
    return dummy_head.next;
  }
};

// 将条件移到循环内部
class Solution2 {
 public:
  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode dummy_head;
    ListNode* p = &dummy_head;
    ListNode* p1 = l1;
    ListNode* p2 = l2;

    int n1;
    int n2;
    int sum;
    int carry = 0;
    while (p1 || p2 || carry) {      
      if (p1) {
        n1 = p1->val;
        p1 = p1->next;
      } else {
        n1 = 0;
      }
      if (p2) {
        n2 = p2->val;
        p2 = p2->next;
      } else {
        n2 = 0;
      }
      sum = n1 + n2 + carry;      
      carry = sum / 10;      
      p->next = new ListNode(sum % 10);
      p = p->next;
    }
    return dummy_head.next;
  }
};