Leetcode solution - 02 Add Two Numbers

Leetcode The first 02.Add Two Numbers topic , questions Medium.

One . Subject requirements

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.

The main idea of the topic

Two integers are stored in two linked lists in reverse order . That is, each node of the linked list stores a digit , The first node stores bits , And so on . Sum through these two linked lists .

Two . Their thinking & Code implementation

Since it's summation , Then it's OK to traverse the two linked lists from scratch and add them , Two problems need to be paid attention to in the process of traversal calculation ：

• Carry problem in calculation , In addition, if there is still carry after traversal, you need to create a new node storage

• There is a linked list traversal completed in advance , Then the value of subsequent traversal of the linked list is 0

Although it is medium Difficult questions , But it's not difficult to do , Just think about the calculation in the traversal process . The implementation code is as follows ：

class Solution {
    
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    
        ListNode result = new ListNode(0);
        ListNode tail = result;
		
		//  Carry or not 
        int isopsephy = 0;

        while (l1 != null || l2 != null) {
    

            int value1 = l1 == null ? 0:l1.val;
            int value2 = l2 == null ? 0:l2.val;

            int val = value1 + value2 + isopsephy;

            if (val > 9) {
    
                val = val % 10;
                isopsephy = 1;

            }else {
    
                isopsephy = 0;
            }

            tail.val = val;

            l1 = l1 == null? null: l1.next;
            l2 = l2 == null? null: l2.next;
            //  Jingle  
            if (l1 == null && l2 == null) {
    
                if (isopsephy > 0) {
    
                    tail.next = new ListNode(isopsephy);
                    tail = tail.next;
                }
                break;
            }
            tail.next = new ListNode(0);
            tail = tail.next;

        }

        return result;
    }
}

• Running results

Runtime: 1 ms, faster than 100.00% of Java online submissions for Add Two Numbers.

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