Leetcode-add-two-numbers

题目描述

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
求两个链表的数之和，仍然以链表形式保存。

思路：新建一个ListNode节点，然后赋给temp节点，变量tmp记录计算进位，当l1.next或者l2.next或者temp！=0的时候进入循环，如果l1不为空，说明l1链表有数，数值累加到tmp中，并且l1 = l1.next；接着l2的处理一样，tmp值加l2节点的值，l2往后移一个位置；计算出tmp后，新建节点new ListNode(tmp % 10),让temp.next指向该节点，同时tmp置为tmp/10。 最后返回最初定义的头节点的next节点。

* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if(l1 == null)
return l2;
if(l2 == null)
return l1;
ListNode dummy = new ListNode(0);
ListNode temp = dummy;

int tmp = 0;
while(l1 != null || l2 != null || tmp != 0){
if(l1 != null){
tmp += l1.val;
l1 = l1.next;
}
if(l2 != null){
tmp += l2.val;
l2 = l2.next;
}

temp.next = new ListNode(tmp%10);
temp = temp.next;
tmp /= 10;

}
return dummy.next;
}
}