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.
```/**
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
let sum = 0;
let carry = 0;
while (l1 !== null || l2 !== null) {
sum = carry + (l1 !== null ? l1.val : 0) + (l2 !== null ? l2.val : 0);
carry = sum >= 10 ? 1 : 0;
sum = sum % 10;

temp.next = new ListNode(sum);
temp = temp.next;

if (l1) l1 = l1.next;
if (l2) l2 = l2.next;

if (carry) temp.next = new ListNode(carry);
}
};```

Given a string, find the first non-repeating character in it and return its index. If it doesn’t exist, return -1.

Examples:

```s = "leetcode"
return 0.

s = "loveleetcode"
return 2.
```

Note: You may assume the string contains only lowercase English letters.

Idea: Hashtable

Solution:

```/**
* @param {string} s
* @return {number}
*/
var firstUniqChar = function(s) {
let map = new Map;

for (let i = 0; i < s.length; i++) {
if (map.has(s[i])) {
map.set(s[i], map.get(s[i]) + 1);
} else {
map.set(s[i], 1);
}
}

for (let i = 0; i < s.length; i++) {
if (map.get(s[i]) === 1) return i;
}
return -1;
};```

Problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Input: nums = [2,7,11,15], target = 9

Output: [0,1]

Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Idea:

Hashtable

Solution:

Hashtable / Javascript

Time complexity: O(n)

Space complexity: O(n)

```/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function(nums, target) {
let map = new Map();

for (let i = 0; i < nums.length; i++) {
let diff = target - nums[i];
if (map.has(diff)) {
return [i, map.get(diff)];
} else {
map.set(nums[i],i);
}
}
};```