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
题解:模拟即可。用carries保存进位,当l1和l2一方为空的时候,余下不为空的链表要单独处理。当l1和l2都为空的时候,如果carries不为空,那么要再单独申请一个链表存放carries,代码如下:
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 14 if(l1 == null && l2 == null) 15 return null; 16 17 ListNode kepeler = new ListNode(0); 18 ListNode head = kepeler; 19 int carries = 0; 20 21 while(l1 != null && l2 != null){ 22 int sum = carries + l1.val + l2.val; 23 carries = sum/10; 24 kepeler.next = new ListNode(sum%10); 25 kepeler = kepeler.next; 26 l1 = l1.next; 27 l2 = l2.next; 28 } 29 30 while(l1 != null){ 31 int sum = carries + l1.val; 32 carries = sum/10; 33 kepeler.next = new ListNode(sum%10); 34 l1 = l1.next; 35 kepeler = kepeler.next; 36 } 37 38 while(l2 != null){ 39 int sum = carries + l2.val; 40 carries = sum/10; 41 kepeler.next = new ListNode(sum%10); 42 l2 = l2.next; 43 kepeler = kepeler.next; 44 } 45 46 if(carries != 0) 47 kepeler.next = new ListNode(carries); 48 49 return head.next; 50 } 51 }