445. Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 8 -> 0 -> 7

---

我的想法是 每次把进位存起来 下次循环时再加

比如输入是 (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)

第一轮相加之后 相加的结果是 7->7->0->7

对应还有一个list存哪些位置应该进位 list={1}

也就是第二个7应该进位

但问题是 如果进位之后导致二次进位 就还需要一次循环

比如加出来的结果是 9->9->9->9

list={3}

那么会导致二次进位 会产生list={2} 

下面的看起来比较简洁 既然不让reverse 那就用stack 效果就如同我们做运算的习惯 先加最后一位

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    Stack<Integer> s1 = new Stack<Integer>();
    Stack<Integer> s2 = new Stack<Integer>();
    
    while(l1 != null) {
        s1.push(l1.val);
        l1 = l1.next;
    };
    while(l2 != null) {
        s2.push(l2.val);
        l2 = l2.next;
    }
    
    int sum = 0;
    ListNode list = new ListNode(0);
    while (!s1.empty() || !s2.empty()) {
        if (!s1.empty()) sum += s1.pop();
        if (!s2.empty()) sum += s2.pop();
        list.val = sum % 10;
        ListNode head = new ListNode(sum / 10);
        head.next = list;
        list = head;
        sum /= 10;
    }
    
    return list.val == 0 ? list.next : list;
}