445 Add Two Numbers II
445. Add Two Numbers II
1. Question
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: (7 -> 2 -> 4 ->3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 ->7
2. Implementation
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode p1 = reverse(l1);
ListNode p2 = reverse(l2);
int sum = 0;
ListNode dummy = new ListNode(0);
ListNode curNode = dummy;
while (p1 != null || p2 != null) {
if (p1 != null) {
sum += p1.val;
p1 = p1.next;
}
if (p2 != null) {
sum += p2.val;
p2 = p2.next;
}
curNode.next = new ListNode(sum % 10);
sum /= 10;
curNode = curNode.next;
}
if (sum != 0) {
curNode.next = new ListNode(sum);
}
return reverse(dummy.next);
}
public ListNode reverse (ListNode head) {
ListNode curNode = head, preNode = null, nextNode = null;
while (curNode != null) {
nextNode = curNode.next;
curNode.next = preNode;
preNode = curNode;
curNode = nextNode;
}
return preNode;
}
}
3. Time & Space Complexity
时间复杂度O(m + n), m是第一个list的长度, n是第二个list的长度,空间复杂度O(1)
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