[leetcode][2] Add Two Numbers
阿新 • • 發佈:2018-09-07
plan leet pub repr put any 原來 integer exce
2. 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.
解析
很簡單用鏈表表示數字相加,註意進位就行了
參考答案(自己寫的)
/** * 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) { ListNode head = new ListNode(0); ListNode index = head; int move = 0; while(l1 != null || l2 != null) { int value = move; if (l1 != null) { value += l1.val; l1 = l1.next; } if (l2 != null) { value += l2.val; l2 = l2.next; } if (value >= 10) { move = 1; value = value - 10; } else { move = 0; } ListNode newNode = new ListNode(value); index.next = newNode; index = newNode; } if (move != 0) { index.next = new ListNode(1); } return head.next; } }
註意點原來我用的是value = value%10,這樣效率很低,其實value的值不會超過10,所以直接value = value - 10就行了。
[leetcode][2] Add Two Numbers