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[LeetCode 2] Add Two Numbers
Published Oct 05, 2021
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Problem
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.
Thoughts
- A dummy head helps with the single linked list problem
- We can loop two lists until they are all null, note their lengths may be different
- Carry value is used in the current value calculation
- Note there may be ending extra carry
Code
Typescript
/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
let dummyHead = new ListNode(0);
let cur = dummyHead;
let l1Val, l2Val, tempVal, curVal, carry = 0;
while(l1 !== null || l2 !== null) {
// handle null node value access
l1Val = l1 === null ? 0 : l1.val;
l2Val = l2 === null ? 0 : l2.val;
// calculate value with the carry
tempVal = l1Val + l2Val + carry;
carry = tempVal >= 10 ? 1 : 0;
curVal = carry ? tempVal - 10 : tempVal
cur.next = new ListNode(curVal);
cur = cur.next;
// go to next node if possible
l1 = l1 === null ? l1 : l1.next;
l2 = l2 === null ? l2 : l2.next;
}
// don't forget the ending carry
if(carry) {
cur.next = new ListNode(1)
}
return dummyHead.next;
};