BiruLyu
6/8/2017 - 8:52 PM

## 445. Add Two Numbers II(Recursive).java

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
Deque<Integer> stack1 = new ArrayDeque<Integer>();
Deque<Integer> stack2 = new ArrayDeque<Integer>();

while(l1 != null) {
stack1.push(l1.val);
l1 = l1.next;
}
while(l2 != null) {
stack2.push(l2.val);
l2 = l2.next;
}

int sum = 0;
while(!stack1.isEmpty() || !stack2.isEmpty() ) {
if(!stack1.isEmpty()) sum += stack1.pop();
if(!stack2.isEmpty()) sum += stack2.pop();
ListNode cur = new ListNode(sum / 10);
sum /= 10;
}
}
}``````
``````public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int size1 = getLength(l1);
int size2 = getLength(l2);
// Make sure l1.length >= l2.length
head.next = size1 < size2 ? helper(l2, l1, size2 - size1) : helper(l1, l2, size1 - size2);
// Handle the first digit
}
}
// get length of the list
public int getLength(ListNode l) {
int count = 0;
while(l != null) {
l = l.next;
count++;
}
return count;
}
// offset is the difference of length between l1 and l2
public ListNode helper(ListNode l1, ListNode l2, int offset) {
if (l1 == null) return null;
// check whether l1 becomes the same length as l2
ListNode result = offset == 0 ? new ListNode(l1.val + l2.val) : new ListNode(l1.val);
ListNode post = offset == 0 ? helper(l1.next, l2.next, 0) : helper(l1.next, l2, offset - 1);
// handle carry
if (post != null && post.val > 9) {
result.val += 1;
post.val = post.val % 10;
}
// combine nodes
result.next = post;
return result;
}``````