［LeetCode］Add Two Numbers

解題思路：

1，創建一個preHead，這樣方便處理第一次運算；

2，加一個carry用於保存 進位數；

3，每一位的運算，都包括 l1.val + l2.val + carry

4，每一次運算結果，都通過 除法和求餘運算，得到 carry和當前位的數字

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param {ListNode} l1
    # @param {ListNode} l2
    # @return {ListNode}
    def addTwoNumbers(self, l1, l2):
        dummy = temp = ListNode(0)
        carry = 0
        while l1 or l2 or carry:
            res = 0
            if l1:
                res += l1.val
                l1 = l1.next
            if l2:
                res += l2.val
                l2 = l2.next
            res += carry
            temp.next = ListNode(res % 10)
            temp = temp.next
            carry = res // 10

        return dummy.next

// Java 實現

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        int carry = 0;
        while(l1 != null || l2 != null){
            int x = (l1 != null ? l1.val : 0);
            int y = (l2 != null ? l2.val : 0);

            int ret = x + y + carry;
            p.next = new ListNode(ret%10);
            p = p.next;
            carry = ret/10;

            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }
        if (carry > 0){
            p.next = new ListNode(carry);
        }

        return dummy.next;
    }
}