Add Two Numbers

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.

思路

• 从头节点开始遍历这两个链表，将每一个对应节点进行相加，结果再除以10作为下一位相加的进位，同时记录余数作为本位的结果，一直处理，直到所有的结点都处理完;

• 当两个链表都到末尾时，判断当前进位是否为1，是的话则增加新节点，值为1；否则直接返回结果;

• 当其中一个链表未到末尾时，将新链表尾部next指向此链表的next位置，并判断进位是否为1，为1的话则执行类似第一步的操作，直到进行为0，返回结果.

• 实现代码：

  public class ListNode {
      int val;
      ListNode next;
      ListNode(int x) {
          val = x;
      }
  }
  
  public ListNode addTwoNumbers(ListNode head1, ListNode head2) {
      if(null == head1 || null == head2) {
          return null;
      }
      ListNode head = new ListNode(0); // 建立新链表的头节点
      ListNode result = head;
      int carry = 0; // 进位，值为 1 或 0
      while (head1.next != null && head2.next != null) {
          head1 = head1.next;
          head2 = head2.next;
          int val = (head1.val + head2.val + carry) % 10;
          carry = (head1.val + head2.val + carry) / 10;
          head.next = new ListNode(val);
          head = head.next;
      }
      if (head1.next != null) { // 链表1未结束
          head.next = head1.next;
      } else if (head2.next != null) { // 链表2未结束
          head.next = head2.next;
      }
  
      if (carry == 1) { // 有进位则继续遍历链表
          while (head.next != null) {
              head = head.next;
              int temp = head.val + carry;
              head.val = temp % 10;
              carry = temp / 10;
          }
          if (carry == 1) {
              head.next = new ListNode(1);
          }
      }
      return result;
  }