leetcode-2 Add Two Numbers 计算链表两个对应和的问题



1.问题描述：

You are given two linked lists representing two non-negativenumbers. The digits are stored in reverse order and each of their nodes containa single digit. Add
the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8
http://www.programcreek.com/2012/12/add-two-numbers/

链表的定义：

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/


这里注意考虑多种情况即可：即考虑两个链表长度不一致，遍历完链表仍有进位的情况。

解法一：较为繁琐的分类讨论方法，避免使用

public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if((l1==null)||(l2==null))
{
return null;
}

int temp_val = l1.val + l2.val;
int go = temp_val /10;
ListNode result = new ListNode(temp_val%10);
l1 = l1.next;
l2 = l2.next;
ListNode temp = result;
while((l1!=null)&&(l2!=null))
{
temp_val = l1.val + l2.val + go;
ListNode temp2 = new ListNode(temp_val%10);
temp.next = temp2;
temp = temp2;
l1 = l1.next;
l2 = l2.next;
go = temp_val /10;
}

while(l1!=null)
{
temp_val = l1.val + go;
ListNode temp2 = new ListNode(temp_val%10);
temp.next = temp2;
temp = temp2;
l1 = l1.next;
go = temp_val /10;
}

while(l2!=null)
{
temp_val = l2.val + go;
ListNode temp2 = new ListNode(temp_val%10);
temp.next = temp2;
temp = temp2;
l2 = l2.next;
go = temp_val /10;
}

if(go != 0)
{
temp.next = new ListNode(go);
}
return result;
}
}


解法二：较优的解法，但在leetcode上运行时间要劣于前者
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry = 0;//表示进位
ListNode head = new ListNode(0);//首前节点，参照C++的尾后节点
ListNode temp = head;
//一直循环至将两个链表遍历完全才退出
while((l1 != null)||(l2 != null))
{
//仅需确定和的几个因子大小即可；
int first = (l1 != null) ? l1.val : 0;//使用if语句亦可
int second = (l2 != null) ? l2.val : 0;

int result = first + second + carry;//每一次循环计算结果
carry = result /10;
ListNode pNode = new ListNode(result % 10);
temp.next = pNode;
temp = pNode;//准备下一循环

if (l1 != null) {
l1 = l1.next;
}

if (l2 != null) {
l2 = l2.next;
}

}
//还需考虑carray不等于0及仍有进位的情况
if (carry > 0)
{
temp.next = new ListNode(carry);
}

return head.next;//注意此返回值
}
}