You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:
Input: (6 -> 2 -> 4 -> 3) + (5 -> 9 -> 4)
Output: 6 -> 8 -> 3 -> 7
import java.util.Stack;
class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
}
}
public class Main {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
Stack<Integer> s1 = new Stack<Integer>();
Stack<Integer> s2 = new Stack<Integer>();
while (l1 != null) {
s1.push(l1.val);
l1 = l1.next;
}
while (l2 != null) {
s2.push(l2.val);
l2 = l2.next;
}
int sum = 0;
ListNode cur = new ListNode(0);
while (!s1.empty() || !s2.empty()) {
if (!s1.empty()) sum += s1.pop();
if (!s2.empty()) sum += s2.pop();
cur.val = sum % 10;
//为啥是“sum/10”，比如输出的[5],[5],输出应该是[1,0];如果是new ListNode(0)则输出是[0]，则错误
ListNode head = new ListNode(sum/10);//先赋值为sum/10，后面cur.val = sum % 10;会赋值
head.next = cur;//新的结点的next指针指向原有的链表头指针
cur = head;//当前指针指向头指针，表示cur指向链表头部
sum /= 10;//用来进位
}
return cur.val == 0 ? cur.next : cur;
}//addTwoNumbers
public static void main(String[] args) {
// Input: (6 -> 2 -> 4 -> 3) + (5 -> 9 -> 4)
// Output: 6 -> 8 -> 3 -> 7
ListNode A = new ListNode(6);
A.next = new ListNode(2);
A.next.next = new ListNode(4);
A.next.next.next = new ListNode(3);
ListNode B = new ListNode(5);
B.next = new ListNode(9);
B.next.next = new ListNode(4);
Main main = new Main();
ListNode C = main.addTwoNumbers(A, B);
System.out.println(C);
}
}
1563 / 1563 test cases passed.
Status: Accepted
Runtime: 63 ms
Your runtime beats 46.11 % of java submissions.
T(n) = O(n)
