題目
2. Add Two Numbers
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
第一次提交
第一次提交代碼：
/**
* 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) {
ListNode pointer1 = l1;
ListNode pointer2 = l2;
while(pointer1 != null && pointer2 != null)
{
int tmp = pointer1.val + pointer2.val;
if(tmp > 9)
{
int x = tmp / 10;
int y = tmp % 10;
pointer1.val = pointer2.val = y;
if(pointer1.next == null && pointer2.next == null)
{
pointer1.next = new ListNode(x);
pointer2.next = new ListNode(0);
}else if(pointer1.next == null)
{
pointer1.next = new ListNode(x);
}else if(pointer2.next == null)
{
pointer2.next = new ListNode(x);
}else
pointer1.next.val += x;
} else
{
pointer1.val = pointer2.val = tmp;
}
pointer1 = pointer1.next;
pointer2 = pointer2.next;
}
if(pointer1 == null)
{
return l2;
} else
{
return l1;
}
}
}
第一次結果細節（圖）：
第一次提交總結：
考慮到要返回同樣的結構的鏈表，故每一位計算後直接同時賦值給l1，l2，返回值時，哪個鏈表還有值就返回哪個。再判斷進位時比較次數過多，還有優化的空間。
第二次提交
第二次提交代碼：
/**
* 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) {
ListNode pointer1 = l1;
ListNode pointer2 = l2;
ListNode pointer1Tmp = l1;
while(pointer1 != null && pointer2 != null)
{
int tmp = pointer1.val + pointer2.val;
if(tmp > 9)
{
int x = tmp / 10;
pointer1.val = tmp % 10;
if(pointer1.next == null)
pointer1.next = new ListNode(x);
else if(pointer2.next ==null)
pointer2.next = new ListNode(x);
else
pointer1.next.val +=x;
} else
{
pointer1.val = pointer2.val = tmp;
}
pointer1Tmp = pointer1;
pointer1 = pointer1.next;
pointer2 = pointer2.next;
}
if(pointer1 == null)
{
pointer1Tmp.next = pointer2;
}
return l1;
}
}
第二次結果細節（圖）：
第二次提交總結：
效率毫無提升！！！
