[leetcode]445. Add Two Numbers II 两数相加II

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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

题目

俩数用链表表示，做个加法

 

思路

1. use stack to covert input list data

2. sum up from pop item from two stacks to a desired linkedlist

 

 

代码

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null && l2 == null) return null;
        if (l2 == null) return l1;
        if (l1 == null) return l2;
        // use stack to convert list's order
        Stack<Integer> s1 = new Stack<>();
        Stack<Integer> s2 = new Stack<>();
    
        while (l1 != null) {
            s1.push(l1.val);
            l1 = l1.next;
        }
        
        while (l2 != null) {
            s2.push(l2.val);
            l2 = l2.next;
        }
        
        int sum = 0;
        ListNode head = new ListNode(0);
        // sum up to a new linkedlist
        while (!s1.isEmpty() || !s2.isEmpty()) {
            if (!s1.isEmpty()) sum += s1.pop();
            if (!s2.isEmpty()) sum += s2.pop();
            
            int val = sum % 10;
          
            ListNode node = new ListNode(0);
            head.val = val;
            node.next = head;
            head = node;
            
            sum /= 10;
        }
        if (sum == 0){
            return head.next;
        }
        head.val = sum;
        return head;
    }
}