268.Missing Number
268. Missing Number
- Given an array
numscontainingndistinct 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.
Method 1
【O(n) time | O(1) space】
package Leetcode.BitManipulation;
/**
* @author zhengstars
* @date 2024/06/11
*/
public class MissingNumber {
public static int findMissingNumber(int[] nums) {
int n = nums.length;
// Calculate the expected sum of numbers from 0 to n using the formula
int totalSum = n * (n + 1) / 2;
// Calculate the sum of elements in the given array
int arraySum = 0;
for (int num : nums) {
arraySum += num;
}
// The missing number is the difference between the totalSum and the arraySum
return totalSum - arraySum;
}
public static void main(String[] args) {
// Test case 1
int[] nums1 = {3, 0, 1};
// Expected output: 2
System.out.println("Missing number: " + findMissingNumber(nums1));
// Test case 2
int[] nums2 = {0, 1};
// Expected output: 2
System.out.println("Missing number: " + findMissingNumber(nums2));
// Test case 3
int[] nums3 = {9, 6, 4, 2, 3, 5, 7, 0, 1};
// Expected output: 8
System.out.println("Missing number: " + findMissingNumber(nums3));
}
}
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