LeeCode是一个有意思的编程网站，主要考察程序员的算法
第二题：
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.
算是一个中度难度的编程题，刚开始考虑转化为整数来加和，但任何类型的整数其实都是有上限的，所以此方法舍弃。
最后自己提交的source如下，想法也是中规中矩吧
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { if(l1 == null || l2 == null) { return null; } // 为了不影响传入参数，设定游标 ListNode cursor1 = l1; ListNode cursor2 = l2; // 结果 ListNode result = new ListNode(0); ListNode cursorResult = result; // 进位 int carry = 0; // 每位加和 int sum; // 暂存游标的值 int cursorVal1 = cursor1.val; int cursorVal2 = cursor2.val; do{ sum = cursorVal1 + cursorVal2 + carry; // 是否有进位 if(sum >= 10) { carry = 1; } else { carry = 0; } cursorResult.next = new ListNode(sum % 10); cursorResult = cursorResult.next; // 如果其中一个游标为空，也就是是说有链表已经遍历完 if(cursor1 != null && cursor1.next != null) { cursor1 = cursor1.next; cursorVal1 = cursor1.val; } else { cursor1 = null; cursorVal1 = 0; } if(cursor2 != null && cursor2.next != null) { cursor2 = cursor2.next; cursorVal2 = cursor2.val; } else { cursor2 = null; cursorVal2 = 0; } } while (cursor1 != null || cursor2 != null); // 最后一个进位为算进去的话 if(carry == 1) { cursorResult.next = new ListNode(1); } // 去掉首个元素 return result.next; } }
提交后性能表现结果如下：
虽然击败了93%的人，但是还有7%的人性能优于我的，看了下官方给出的答案，整体思路是一样的，而且它将x,y的声明放在循环内部。这可能也是统计的不精确吧
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode dummyHead = new ListNode(0); ListNode p = l1, q = l2, curr = dummyHead; int carry = 0; while (p != null || q != null) { int x = (p != null) ? p.val : 0; int y = (q != null) ? q.val : 0; int sum = carry + x + y; carry = sum / 10; curr.next = new ListNode(sum % 10); curr = curr.next; if (p != null) p = p.next; if (q != null) q = q.next; } if (carry > 0) { curr.next = new ListNode(carry); } return dummyHead.next; }
时间复杂度和空间复杂度如下：
Complexity Analysis
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Time complexity : O(\max(m, n))O(max(m,n)). Assume that mm and nn represents the length of l1l1 and l2l2 respectively, the algorithm above iterates at most \max(m, n)max(m,n) times.
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Space complexity : O(\max(m, n))O(max(m,n)). The length of the new list is at most \max(m,n) + 1max(m,n)+1.
参考：
https://leetcode.com/problems/add-two-numbers
