https://leetcode.com/problems/add-two-numbers-ii/description/
445. Add Two Numbers II
給你兩個鏈表l1和l2，返回他們的相加（十進位加法），題目保證 l1 和 l2 不會是 0
開頭的數字（除了0）。
Example 1:
https://assets.leetcode.com/uploads/2021/04/09/sumii-linked-list.jpg
Input: l1 = [7,2,4,3], l2 = [5,6,4]
Output: [7,8,0,7]
Example 2:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [8,0,7]
Example 3:
Input: l1 = [0], l2 = [0]
Output: [0]
思路：
1. 因為十進位加法要從右加到左(因為會產生進位)，所以我們把鏈表 l1 和 l2 做反轉。
2. 將 l1 和 l2 做加法運算產生一個新的鏈表 l3。
3. 在得到解後需再將鏈表 l3 反轉，因為我們鏈表的連結方向和相加結果相反。
Java Code：
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class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
l1 = reverseList(l1);
l2 = reverseList(l2);
ListNode l3 = mergeList(l1, l2);
return reverseList(l3);
}
private ListNode reverseList(ListNode curr) {
ListNode prev = null;
while (curr != null) {
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
private ListNode mergeList(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
ListNode curr = dummy;
int carry = 0;
while (l1 != null || l2 != null) {
int sum = carry;
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}
curr.next = new ListNode(sum % 10);
curr = curr.next;
carry = sum > 9 ? 1 : 0;
}
if (carry != 0) {
curr.next = new ListNode(1);
}
return dummy.next;
}
}
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