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# 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:
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609092186@qq.com或者留言，我尽快回复）
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原文地址：http://www.cnblogs.com/Piers/p/5739833.html
