2021/04/26
如果允许修改链表,就分别反转后相加,再反转。
题目不允许修改链表,就是用栈来做。
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
LinkedList<Integer> stack1 = new LinkedList<>();
LinkedList<Integer> stack2 = new LinkedList<>();
while (l1 != null) {
stack1.push(l1.val);
l1 = l1.next;
}
while (l2 != null) {
stack2.push(l2.val);
l2 = l2.next;
}
ListNode head = null;
int c = 0;
while (!stack1.isEmpty() || !stack2.isEmpty() || c != 0) {
int a = stack1.isEmpty() ? 0 : stack1.pop();
int b = stack2.isEmpty() ? 0 : stack2.pop();
int s = (a + b + c) % 10;
c = (a + b + c) / 10;
ListNode node = new ListNode(s);
node.next = head;
head = node;
}
return head;
}
}