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
Output: 7 -> 0 -> 8
Example
Given
7->1->6 + 5->9->2. That is, 617 + 295.
Return
2->1->9. That is 912.
Given
3->1->5 and 5->9->2, return 8->0->8.
维护一个carry用于进位 因为是反过来写的所以进位在后面 可以直接加
时间复杂度O(n)
时间复杂度O(n)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
int carry = 0;
ListNode head = new ListNode(-1);
ListNode l3 = head;
while (l1 != null || l2 != null) {
if (l1 != null) {
carry += l1.val;
l1 = l1.next;
}
if (l2 != null) {
carry += l2.val;
l2 = l2.next;
}
l3.next = new ListNode(carry%10);
carry = carry/10;
l3 = l3.next;
}
if (carry > 0) {
l3.next = new ListNode(carry);
}
return head.next;
}
}
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