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.

<解題>SUM

class Solution {
    public int missingNumber(int[] nums) {
        int sum1 = 0;
        int sum2= 0;
        
        for(int num : nums){
            sum1 += num ;
        }
        for (int i = 0; i<= nums.length; i++){
            sum2 += i;
        }
        return sum2-sum1;
    }
}

Time: O(n) Space: O(1)

<解題>XOR

這題也可以用XOR，利用i和nums[i]找到一樣的數字就抵銷，剩下的就是missing number

public int missingNumber(int[] nums) {
    int res = nums.length;
    for(int i=0; i<nums.length; i++){
        res ^= i;
        res ^= nums[i];
    }
    return res;
}

Time: O(n) Space: O(1)