🖥️ 문제
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:
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:
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.
📝 풀이
class Solution {
public int missingNumber(int[] nums) {
int n = nums.length;
int sum = (n * (n + 1)) / 2;
int result = 0;
for (int i = 0; i < nums.length; i++) {
result = result + nums[i];
}
return sum - result;
}
}
문제는 o부터 n까지의 인덱스에서 빠진 숫자를 찾는 것이다. 등차수열을 이용해 0부터 n까지의 숫자의 합을 구한 뒤, for문을 활용해서 배열의 숫자들을 하나씩 빼니 쉽게 구할 수 있었다.
