Question

Source

• leetcode: Add Two Numbers | LeetCode OJ
• lintcode: Add Two Numbers

Problem

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

Example

Given 7->1->6 + 5->9->2 . That is, 617 + 295 .

Return 2->1->9 . That is 912 .

Given 3->1->5 and 5->9->2 , return 8->0->8 .

题解

一道看似简单的进位加法题，实则杀机重重，不信你不看答案自己先做做看。

首先由十进制加法可知应该注意进位的处理，但是这道题仅注意到这点就够了吗？还不够！因为两个链表长度有可能不等长！因此这道题的亮点在于边界和异常条件的处理，感谢 @wen 引入的 dummy 节点，处理起来更为优雅！

Python

# Definition for singly-linked list.
# class ListNode(object):
#   def __init__(self, x):
#     self.val = x
#     self.next = None

class Solution:
  def add_two_numbers(self, l1, l2):
    '''
    :type l1: ListNode
    :type l2: ListNode
    :rtype: ListNode
    '''
    carry = 0
    dummy = prev = ListNode(-1)
    while l1 or l2 or carry:
      v1 = l1.val if l1 else 0
      v2 = l2.val if l2 else 0
      val = (v1 + v2 + carry) % 10
      carry = (v1 + v2 + carry) / 10

      prev.next = ListNode(val)
      prev = prev.next

      if l1:
        l1 = l1.next
      if l2:
        l2 = l2.next
    return dummy.next

C++

/**
 * 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 dummy(0);
    ListNode *curr = &dummy;
    int carry = 0;

    while ((l1 != NULL) || (l2 != NULL) || (carry != 0)) {
      int l1_val = (l1 != NULL) ? l1->val : 0;
      int l2_val = (l2 != NULL) ? l2->val : 0;
      int sum = carry + l1_val + l2_val;
      carry = sum / 10;
      curr->next = new ListNode(sum % 10);

      curr = curr->next;
      if (l1 != NULL) l1 = l1->next;
      if (l2 != NULL) l2 = l2->next;
    }

    return dummy.next;
  }
};

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) { val = x; }
 * }
 */
public class Solution {
  public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode curr = dummy;
    int carry = 0;

    while ((l1 != null) || (l2 != null) || (carry != 0)) {
      int l1_val = (l1 != null) ? l1.val : 0;
      int l2_val = (l2 != null) ? l2.val : 0;
      int sum = carry + l1_val + l2_val;
    // update carry
      carry = sum / 10;
      curr.next = new ListNode(sum % 10);

      curr = curr.next;
      if (l1 != null) l1 = l1.next;
      if (l2 != null) l2 = l2.next;
    }

    return dummy.next;
  }
}

源码分析

1. 迭代能正常进行的条件为 (NULL != l1) || (NULL != l2) || (0 != carry) , 缺一不可。
2. 对于空指针节点的处理可以用相对优雅的方式处理 - int l1_val = (NULL == l1) ? 0 : l1->val;
3. 生成新节点时需要先判断迭代终止条件 - (NULL == l1) && (NULL == l2) && (0 == carry) , 避免多生成一位数 0。 使用 dummy 节点可避免这一情况。

复杂度分析

没啥好分析的，时间和空间复杂度均为 O(max(L1,L2))O(max(L1, L2))O(max(L1,L2)).

Reference

• CC150 Chapter 9.2 题 2.5，中文版 p123
• Add two numbers represented by linked lists | Set 1 - GeeksforGeeks