Javascript in Leetcode 01–03 | 面試問題
      Pheezx Coding
01. Two sum
Input: nums = [2,7,11,15], target = 9 Output: [0,1] Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
solution
function twoSum(nums, target) {
 const comp = new Map();
  const len = nums.length;
    
    for(let i=0; i < len; i++){
        if(comp[nums[i]] >= 0){
            return [comp[nums[i]], i];
        }
        comp[target- nums[i]] = i;
    }
    return [];
}
const nums = [2,7,11,15];
console.log(twoSum(nums,9));
02. Add two numbers
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.
Input: l1 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807.
Solution
var addTwoNumbers = function(l1, l2) {
    let carry = 0;
    let previousNode = new ListNode();
    const headNode = previousNode;
    while(l1 || l2 || carry) {
        let val1 = 0;
        let val2=0;
        if(l1) {
            val1=l1.val;
            l1=l1.next;
        }
        if(l2) {
            val2=l2.val;
            l2=l2.next;
        }
        const sum = val1 + val2 + carry;
        carry = Math.floor(sum / 10);
        const digit = sum % 10;
        const currentNode =  new ListNode(digit);
        previousNode.next = currentNode;
        previousNode = currentNode;
    }
    return headNode.next;
};
leetcode problems | https://leetcode.com/problems/add-two-numbers/
03. Longest Substring Without Repeating Characters
Input: s = "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3.
Constraints:
0 <= s.length <= 5 * 104sconsists of English letters, digits, symbols and spaces.
Solution
var lengthOfLongestSubstring = function(s) {
    let max = 0
    let begin = 0
    let set = new Set()
    
    for(let end = 0; end < s.length; end++){
        while(set.has(s[end])){
            set.delete(s[begin])
            begin = begin + 1
        }
            set.add(s[end])
            max = Math.max(max, end - begin + 1)
    }
    return max
        
};
