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.
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.
Constraints:
n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
All the numbers of nums are unique.
var missingNumber = function (nums) {
let num = nums.length;
for (let i = 0; i <= num; i++) {
if (!nums.includes(i)) {
return i;
}
}
};
✅ 0부터 num까지 반복문을 돌리면서 없는 번호만 리턴하면 되는 문제