leetcode-2-Add Two Numbers

leetcode-2-Add Two Numbers

Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

common

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/**
* 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 dummyHead = new ListNode(0); // 一点要赋值一个节点,进行操作
ListNode p = l1, q = l2, curr = dummyHead;
int carry = 0;
while (p != null || q != null) {
int x = (p != null) ? p.val : 0;
int y = (q != null) ? q.val : 0;
int sum = carry + x + y;
carry = sum / 10;
curr.next = new ListNode(sum % 10);
curr = curr.next;
if (p != null) p = p.next;
if (q != null) q = q.next;
}
if (carry > 0) {
curr.next = new ListNode(carry);
}
return dummyHead.next;
}
}

best

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/**
* 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 head = new ListNode(0);
int carry = 0;
while(l1!=null||l2!=null||carry>0)
{
ListNode itr = head;
while(itr.next!=null)
itr = itr.next; // 寻找最后一个节点
int sum = ( (l1==null ? 0 : l1.val) + (l2==null ? 0 : l2.val) + carry);
carry = sum/10;
ListNode temp = new ListNode(sum%10);
itr.next = temp;
if(l1!=null)
l1 = l1.next;
if(l2!=null)
l2 = l2.next;
}

return head.next;
}
}

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