268.missing number of leetcode

1. Description

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.

2. Follow Up

Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

3. Test Cases

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.

Example 4:

Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.

4. Solution

4.1

If you don't think of any strange tricks , Then write an answer honestly first .

class Solution {
    
    public int missingNumber(int[] nums) {
    
        Set<Integer> numSet = new HashSet<Integer>();
        
        for(int num : nums) {
    
            numSet.add(num);
        }
        
        int expectedNumCount = nums.length + 1;
        
        for (int number=0; number<expectedNumCount; number++) {
    
            if (!numSet.contains(number)) {
    
                return number;
            }
        }
        
        return -1;
    }
}

The efficiency of this is obviously not high .

4.3

Here you can use special skills .
XOR operation is a magical operation in computer , It can be simply understood as the same as 0, Different for 1. If you use 0 And any number XOR , Then the result is any number .
here , Because we know that only one number is missing in the array , And the size of the numerical value is the length of the array +1 The size of is the maximum . Then we can see the array [3,0,1], The array length is 3, The corresponding number is 0,1,2,3 Four Numbers , But one is missing . Consider the corresponding subscript , Respectively 0,1,2, In this way, we can use the subscript of the array and the exclusive or of the array word , More than the length of the array +1, namely
3 ^ 0 ^ 3 ^ 1 ^ 0 ^ 2 ^ 1 = 2, That is, what is missing is 2.

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