2154. Keep Multiplying Found Values by Two

Easy

題目描述

Description

You are given an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums.

You then do the following steps:

• If original is found in nums, multiply it by two (i.e., set original = 2 * original).
• Otherwise, stop the process.
• Repeat this process with the new number as long as you keep finding the number.

Return the final value of original.

Example 1:

Input: nums = [5,3,6,1,12], original = 3
Output: 24
Explanation:

• 3 is found in nums. 3 is multiplied by 2 to obtain 6.
• 6 is found in nums. 6 is multiplied by 2 to obtain 12.
• 12 is found in nums. 12 is multiplied by 2 to obtain 24.
• 24 is not found in nums. Thus, 24 is returned.

Example 2:

Input: nums = [2,7,9], original = 4
Output: 4
Explanation:

• 4 is not found in nums. Thus, 4 is returned.

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i], original <= 1000

解答

Solution

/**
 * @param {number[]} nums
 * @param {number} original
 * @return {number}
 */
var findFinalValue = function (nums, original) {
  while (nums.includes(original)) {
    original *= 2;
  }

  return original;
};

解題思路

題目要求的是在 nums 中找到 original，找到後要將原本的 original 乘 2，然後使用新的 original 再次尋找。
所以可以先定義一個 while 迴圈，如果 nums 中包含 original 就重複執行，直到找不到 original 就跳出迴圈並 return 答案。

var findFinalValue = function (nums, original) {
  while (nums.includes(original)) {
    original *= 2;
  }

  return original;
};

心得

把題目描述轉換成程式碼