# 题目
2. Add Two Numbers
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
# 思路
不知道各位能不能看懂题目，简单解释一下，就是把整数每一位颠倒进行加法。题目中给出的例子，最初对应342 + 465 = 807，然后颠倒变成243 + 564 = 708，在转换为链表。
这下题目给出链表的定义，我们需要对这种类型的链表进行操作。
// Definition for singly-linked list.
public class ListNode
{
public int val;
public ListNode next;
public ListNode(int x) { val = x; }
}
方法一：普通遍历，链表l1和l2相应位置相加，再加进位，存入链表result中。
注意点:
- 对于一段长于另外一段的链表部分，单独处理。
- 进位。
- 结果之和长于两个链表的情况，如1 + 999 = 1000。
普通遍历，时间复杂度O(n)，空间复杂度O(n)，时间204ms。
// normal traversal: time O(n) space O(n) result: 204ms
public void calculateSum(ListNode tresult, ref int carry, int sum)
{
if (sum >= 10)
{
carry = 1;
tresult.next = new ListNode(sum - 10);
}
else
{
carry = 0;
tresult.next = new ListNode(sum);
}
}
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
ListNode tl1 = l1, tl2 = l2;
ListNode result = new ListNode(0);
ListNode tresult = result;
int carry = 0;
// both ListNode 1 and ListNode 2 have values
while (tl1 != null && tl2 != null)
{
calculateSum(tresult, ref carry, tl1.val + tl2.val + carry);
tl1 = tl1.next;
tl2 = tl2.next;
tresult = tresult.next;
}
// Debug.Assert(!(tl1 != null && tl2 != null), "tl1 and tl2 aren‘t null");
// either ListNode 1 or ListNode 2 has values (maybe) and don‘t forget carry.
while (tl1 != null)
{
calculateSum(tresult, ref carry, tl1.val + carry);
tl1 = tl1.next;
tresult = tresult.next;
}
while (tl2 != null)
{
calculateSum(tresult,ref carry, tl2.val + carry);
tl2 = tl2.next;
tresult = tresult.next;
}
// at this time, ListNode 1 and ListNode 2 should be null, however, carry could be null or not
// Debug.Assert(tl1 == null && tl2 == null, "calculation doesn‘t finish");
if (carry == 1) tresult.next = new ListNode(1);
// neither ListNode 1 nor ListNode 2 have values
return result.next;
}
*/
方法二：空间优化遍历，链表l1和l2相应位置相加，再加进位，存入链表l1中。方法二的代码没有方法一的代码清晰。
空间优化遍历，时间复杂度O(n)，空间复杂度O(1)，时间208ms。
// use ListNode 1 to restore result
// space (and time, I think, but result doesn‘t prove) optimized traversal: time O(n) space O(1) result: 208ms
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
if (l1 == null) return l2;
if (l2 == null) return l1;
int carry = 0, sum = 0;
ListNode pre = null, result = l1;
while (l1 != null || l2 != null || carry != 0)
{
// calculate sum and carry
sum = 0;
if (l1 != null) sum += l1.val;
if (l2 != null)
{
sum += l2.val;
l2 = l2.next; // ListNode1 will be used below, ListNode2 not, so if ListNode 2 next exists, ListNode 2 move to next
}
sum += carry;
if (sum >= 10)
{
carry = 1;
sum -= 10;
}
else
{
carry = 0;
}
// find a place for sum in ListNode 1, l1 is in use
if (l1 != null)
{
pre = l1;
if (sum >= 10) sum -= 10;
l1.val = sum;
l1 = l1.next;
}
else
{
if (sum >= 10) sum -= 10;
pre.next = new ListNode(sum);
pre = pre.next;
}
}
return result;
}
*/
方法三：递归，链表l1和l2相应位置相加，再加进位，存入链表node中。速度最快，是比较好的解决方案。
# 解决（递归）
递归，时间复杂度O(n)，空间复杂度O(n)，时间196ms。
// recursive tranversal: time O(n) space:O(n) time: 196ms (why it is faster than normal loop)
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
return AddTwoNumbers(l1, l2, 0);
}
public ListNode AddTwoNumbers(ListNode l1, ListNode l2, int carry)
{
if (l1 == null && l2 == null && carry == 0) return null;
// calculate sum
int sum = 0;
if (l1 != null) sum += l1.val;
if (l2 != null) sum += l2.val;
sum += carry;
if (sum >= 10)
{
carry = 1;
sum -= 10;
}
else
{
carry = 0;
}
// set node‘s next and val and return
ListNode node = new ListNode(sum);
node.next = AddTwoNumbers(l1 != null ? l1.next : null, l2 != null ? l2.next : null, carry);
return node;
}
# 题外话
为何递归会比循环快呢？百思不得其解，若有高人知道，请指教。
# 测试用例
static void Main(string[] args)
{
_2AddTwoNumbers solution = new _2AddTwoNumbers();
ListNode l1 = new ListNode(1);
ListNode l2 = new ListNode(2);
ListNode result = new ListNode(3);
// ListNode doesn‘t have a null constructor, so we can igonore this case
Debug.Assert(Test.match(solution.AddTwoNumbers(l1, l2), result), "wrong 1");
// ListNode 1 length is larger than ListNode 2 length
l1 = new ListNode(1);
l1.next = new ListNode(4);
l1.next.next = new ListNode(5);
l2 = new ListNode(2);
result.next = new ListNode(4);
result.next.next = new ListNode(5);
Debug.Assert(Test.match(solution.AddTwoNumbers(l1, l2), result), "wrong 2");
// ListNode 2 length is larger than ListNode 1 length and has carries
l1 = new ListNode(1);
l1.next = new ListNode(1);
l2 = new ListNode(9);
l2.next = new ListNode(8);
l2.next.next = new ListNode(9);
result = new ListNode(0);
result.next = new ListNode(0);
result.next.next = new ListNode(0);
result.next.next.next = new ListNode(1);
Debug.Assert(Test.match(solution.AddTwoNumbers(l1, l2), result), "wrong 3");
}
class Test
{
public static bool match(_2AddTwoNumbers.ListNode l1, _2AddTwoNumbers.ListNode l2)
{
_2AddTwoNumbers.ListNode tl1 = l1, tl2 = l2;
while(tl1 != null && tl2 != null)
{
if (tl1.val != tl2.val) return false;
tl1 = tl1.next;
tl2 = tl2.next;
}
if (tl1 == null && tl2 == null) return true;
else return false;
}
}
# 地址
Q： https://leetcode.com/problems/add-two-numbers/
A： https://github.com/mofadeyunduo/LeetCode/tree/master/2AddTwoNumbers
（希望各位多多支持本人刚刚建立的GitHub和博客，谢谢，有问题可以邮件[email protected]或者留言，我尽快回复）
时间： 2024-10-01 07:02:49