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.
题目:
用两个链表表示的两个数, 求相加之和
Solution1: For given two lists, keep looping where either one isn't null. When ListNode comes to null, consider its val as 0.
For result list, use a dummy node.
- code:
- /**
- * Definition for singly-linked list.
- * public class ListNode {
- * int val;
- * ListNode next;
- * ListNode(int x) { val = x; }
- * }
- */
- /*
- Time Complexity: O(max(l1,l2))
- Space Complexity: O(max(l1,l2))
- */
- class Solution {
- public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
- ListNode dummy = new ListNode(-1);
- ListNode pre = dummy;
- ListNode p1 = l1;
- ListNode p2 = l2;
- int carry = 0;
- while(p1 != null || p2 != null){
- int x = (p1 != null) ? p1.val : 0;
- int y = (p2 != null) ? p2.val : 0;
- int sum = carry + x + y;
- carry = sum / 10;
- // generate the result list
- pre.next = new ListNode(sum % 10);
- // move three lists pointers
- pre = pre.next;
- if(p1 != null) p1 = p1.next;
- if(p2 != null) p2 = p2.next;
- }
- // in case last two digit sum> 10
- if(carry> 0){
- pre.next = new ListNode(carry);
- }
- return dummy.next;
- }
- }
[leetcode]2. Add Two Numbers 两数相加
来源: http://www.bubuko.com/infodetail-3009542.html