leetcode-不同类型的数相加

好吧到今天开始我就完成了leetcode第二个主题，每天一道编程题必须要坚持下去，这次我做的题大多是让两个东西相加，这个东西可以是两个整型字符串、二进制字符串、整型链表等等。
我个人觉得这道题的灵感主要是来自《数字逻辑》里面的全加器，这个全加器大约是这个样子的

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A和B是两个本位数Cin是来自低位的进位，S是面向高位的进位，基本上依靠这个就能把这类题的最主要的大体循环结构搞出来了：
示例代码

//假设传入的参数是两个整型字符串num1和num2
StringBuilder stringBuilder = new StringBuilder();
for (int i = num1.length() - 1, j = num2.length() - 1; i >= 0 || j >= 0; ) {
            int sum = carry;
            if (i >= 0) sum += num1.charAt(i--) - '0';
            if (j >= 0) sum += num2.charAt(j--) - '0';
            stringBuilder.append(sum % 10);
            carry = sum / 10;
        }

来道例题：

2. Add Two Numbers
   You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
然后我的解题方法是：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carry = 0;
        ListNode corrent = new ListNode(-20);
        ListNode result = corrent;
        for (ListNode list1 = l1, list2 = l2; list1 != null || list2 != null || carry > 0; ) {
            int sum = carry;
            if (list1 != null) {
                sum += list1.val;
                list1 = list1.next;
            }
            if (list2 != null) {
                sum += list2.val;
                list2 = list2.next;
            }

            corrent.next = new ListNode(sum % 10);
            corrent = corrent.next;
            carry = sum / 10;
        }
        
        return result.next;
    }

emmm最后的运行情况：

image.png

类似的题还有这些：

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