leetcode#2 Add Two Numbers

Question

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 contains a single digit. Add the two numbers and return the sum as a linked list.

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

 

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0] Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9] Output: [8,9,9,9,0,0,0,1]

 

Constraints:

• The number of nodes in each linked list is in the range [1, 100].
• 0 <= Node.val <= 9
• It is guaranteed that the list represents a number that does not have leading zeros.

 

solution

var addTwoNumbers = function(l1, l2) {
    var result = [];
    
    var l1_len = l1.length;
    var l2_len = l2.length;
    
    var len = l1_len >= l2_len ? l1_len : l2_len;
    
    var sumValue = 0;
    var overCnt = 0;
    
    for(var i=0; i<len; i++) {
        if(l1[i] === undefined) l1[i] = 0;
        if(l2[i] === undefined) l2[i] = 0;

        sumValue = l1[i] + l2[i] + overCnt;
        
        if(sumValue >= 10) {
            sumValue = sumValue % 10;
            overCnt = 1;
            result.push(sumValue);
            
            if (i === len-1) {
                result.push(1);
            }
            
        } else {
            overCnt = 0;
            result.push(sumValue);
        }
    }
    
    return result;
};

 

other solution

var addTwoNumbers = function(l1, l2) {
    let newNode = new ListNode(0);
    let head = newNode
    var carry = false
    while(l1 || l2){
        var nodeSum =  (l1?.val || 0) +  (l2?.val || 0);
        nodeSum  += (carry) ? 1 : 0;
        if(nodeSum > 9){
            carry = true
        }else{
            carry = false
        }
        head.next = new ListNode(nodeSum % 10)
        
        if(l1)l1 = l1.next;
        if(l2)l2 = l2.next;
        
        head = head.next
    }
    if(carry){
         head.next = new ListNode(carry);
    }
    return newNode.next
};