LeetCode Js-268. Missing Number

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

給予一個陣列 nums 包含 n 的數字在 [0, n] 的範圍，回傳數組中唯一缺少的數字。

Example 1:

Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

Example 2:

Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

Example 3:

Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

Solution:
1. 宣告 result 為 0。
2. 宣告 numLength 為 nums 的長度。
3. 如果沒有長度，返回 0。
4. 執行 nums 長度的迴圈，且每一輪都將 result 加上 nums 陣列中的數字，
同時減去當前第 i 輪的 i 值。
(正常數列的加總 – 減去該陣列實際值得加總 = 缺少的數字)
5. 回傳 nums 的長度減去 result。

Code 1: BigO(n)

var missingNumber = function(nums) {
  let length = nums.length,
      numsTotal = 0

  if (length === 0) return 0;

  for (let i = 0; i < length; i++) {
    numsTotal += nums[i] - i
  }

  return length - numsTotal;
};

Input: nums = [3,0,1]

numsTotal += nums[i] - i
numsTotal = 0 + 3 - 0 = 3
numsTotal = 3 + 0 - 1 = 2
numsTotal = 2 + 1 - 2 = 1

return length - numsTotal = 3 - 1 = 2

Input: nums = [0,1]

numsTotal += nums[i] - i
numsTotal = 0 + 0 - 0 = 0
numsTotal = 0 + 1 - 1 = 0

return length - numsTotal = 2 - 0 = 2

Input: nums = [9,6,4,2,3,5,7,0,1]

numsTotal += nums[i] - i
numsTotal = 0 + 9 - 0 = 9
numsTotal = 9 + 6 - 1 = 14
numsTotal = 14 + 4 - 2 = 16
numsTotal = 16 + 2 - 3 = 15
numsTotal = 15 + 3 - 4 = 14
numsTotal = 14 + 5 - 5 = 14
numsTotal = 14 + 7 - 6 = 15
numsTotal = 15 + 0 - 7 = 8
numsTotal = 8 + 1 - 8 = 1

return length - numsTotal = 9 - 1 = 8

var missingNumber = function(nums) {
  let length = nums.length
      lengthTotal = 0,
      numsTotal = 0
  
  if (length === 0) return 0;

  for (let i = 0; i <= length; i++) {
    lengthTotal += i
  }
  for (let i = 0; i < length; i++) {
    numsTotal += nums[i]
  }
  return lengthTotal - numsTotal;  
};

var missingNumber = function(nums) {
  let numsSet = new Set(nums),
  lostNumbers = []

  for (let i = 0; i <= nums.length; i++) {
    if (!numsSet.has(i)) lostNumbers.push(i)
  }
  return lostNumbers;
};

var missingNumber = function(nums) {
  let numsSet = new Set(nums)

  for (let i = 0; i <= nums.length; i++) {
    if (!numsSet.has(i)) return i;
  }
};