【leetcode】2-Add Two Numbers
阿新 • • 發佈:2018-12-12
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.
/** * 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 temp = new ListNode(-1); ListNode res = temp; int jw = 0; while(l1!=null || l2!=null){ int d1 = l1 == null ? 0 : l1.val; int d2 = l2 == null ? 0 : l2.val; int sum = d1 + d2 + jw; jw = sum >= 10 ? 1 : 0; temp.next = new ListNode(sum % 10); temp = temp.next; if(l1!=null) l1 = l1.next; if(l2!=null) l2 = l2.next; } if(jw == 1){ temp.next = new ListNode(1); } return res.next; } }
* 連結串列題目要設定一個提前點,然後一個新點用來返回結果 * 這道題一開始沒設定提前點,結果最後又new了一個廢點,所以用最後判斷非null的話都向後挪一位變成都是null,所以跳出迴圈結束,就不再new新的temp點 * 題目最後加特殊判斷來看是否需要進位,因為這已經跳出while迴圈了