Article 13 title Add Numbers

You are given, linked lists representing, and non-negative numbers. The digits is stored in reverse order and all of their nodes contain a single digit. ADD the numbers and return it as a linked list.

Input: (2, 4, 3) + (5, 6, 4)

Output: 7, 0, 8

Solution1: Do not allocate extra space, but change the value in the original lists.

/** * Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int x) {* val = x; * Next = Null *} *} */public class Solution {public ListNode addtwonumbers (listnode L1, ListNode L2) {if (l1==n        Ull&&l2==null) return null;        if (l1==null) return L2;        if (l2==null) return L1;        ListNode ptr1 = L1, ptr2 = L2;         int c = 0;            ListNode Prev=l1;//reserve The previous value in the generated list while (Ptr1!=null && ptr2!=null) {            Ptr1.val = Ptr1.val+ptr2.val+c;            c = PTR1.VAL/10;            Ptr1.val = ptr1.val%10;            prev = PTR1;            Ptr1=ptr1.next;        Ptr2=ptr2.next;            } if (ptr2!=null) {//ptr1==null prev.next = ptr2;            PTR1 = PTR2;  Ptr2=null; Must do the maintaining end condition} while (C!=0&&ptr1!=null) {ptr1.Val =ptr1.val+c;            c = PTR1.VAL/10;            Ptr1.val = ptr1.val%10;            prev = PTR1;        PTR1 = Ptr1.next;            } if (c!=0&&ptr1==null&&ptr2==null) {ListNode newNode =new listnode (1);            Prev.next = NewNode;        PTR1 = NewNode;    } return L1; }}

Solution 2: Does not change the value in the original lists. But a new list needs to be created

 * public class ListNode {* int val; * ListNode Next; * ListNode (int x) {* val = x; * Next = Null *} *} */public class Solution {public ListNode addtwonumbers (listnode L1, ListNode L2) {if (l1==n        Ull&&l2==null) return null;        if (l1==null) return L2;        if (l2==null) return L1;        ListNode head = new ListNode (0);         int c = 0; ListNode Prev=head;//reserve The previous value in the generated list while (L1!=null | | l2!=null) {LISTN            Ode cur = new ListNode (0);                if (l1!=null) {cur.val + = L1.val;            L1=l1.next;                } if (L2!=null) {cur.val + = L2.val;            L2=l2.next;            } Cur.val + = C;            c = CUR.VAL/10;            Cur.val = cur.val%10;            Prev.next = cur;        prev = cur;            } if (c!=0) {ListNode newNode =new listnode (1); Prev.next = NewNode;    } return head.next; }}

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Article 13 title Add Numbers