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
Solution1：不用分配多余的空间，可是会改变原先lists中的值。
/**
* 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==null&&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 this for 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: 不会改变原lists中的值。但需建立一个新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==null&&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){
ListNode 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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