[LeetCode] 445. Add Two Numbers II 两个数字相加之二

You are given two linked lists representing two non-negative numbers. 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

2. Add Two Numbers 的变形，之前的题最高位在链表末位，此题链表头部表示高位，尾部表示低位，不允许反转链表。两个数相加需要从低位开始。可以利用Stack的特点后进先出，遍历两个链表，将数字分别压入两个栈s1和s2，然后开始循环，如果栈不为空，则将栈顶数字加入sum中。

Java:

public class Solution {
    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;
    }
}

Python:

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        stk1, stk2 = [], []
        while l1:
            stk1.append(l1.val)
            l1 = l1.next
        while l2:
            stk2.append(l2.val)
            l2 = l2.next
 
        prev, head = None, None
        sum = 0
        while stk1 or stk2:
            sum /= 10
            if stk1:
                sum += stk1.pop()
            if stk2:
                sum += stk2.pop()
             
            head = ListNode(sum % 10)
            head.next = prev
            prev = head
 
        if sum >= 10:
            head = ListNode(sum / 10)
            head.next = prev
 
        return head

C++:

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        stack<int> stk1, stk2;
        while (l1) {
            stk1.emplace(l1->val);
            l1 = l1->next;
        }
        while (l2) {
            stk2.emplace(l2->val);
            l2 = l2->next;
        }
 
        ListNode *prev = nullptr, *head = nullptr;
        int sum = 0;
        while (!stk1.empty() || !stk2.empty()) {
            sum /= 10;
            if (!stk1.empty()) {
                sum += stk1.top();
                stk1.pop();
            }
             
            if (!stk2.empty()) {
                sum += stk2.top();
                stk2.pop();
            }
             
            head = new ListNode(sum % 10);
            head->next = prev;
            prev = head;
        }
 
        if (sum >= 10) {
            head = new ListNode(sum / 10);
            head->next = prev;
        }
         
        return head;
    }
};