0268-missing-number.js

//Blog: https://www.allenliservice.online/leetcode-js-268-missing-number/

// <strong>Solution:</strong>
// 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。

/* <strong>Code 1: BigO(n)</strongc> */
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;
};

/* <strong>FlowChart:</strong>
<strong>Example 1</strong>
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
*/

// <strong>Code 2: BigO(2n)</strongc>
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;
};

/* <strong>Code 3: BigO(n) ES6 Set</strongc> */
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;
};

//<strong>Code 4: BigO(n) ES6 Set</strongc>
var missingNumber = function (nums) {
  let numsSet = new Set(nums);

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