[LeetCode] C++: Simple Questions-Bit Operation 268. Missing Numbers

268. Missing Numbers

Easy difficulty 364

Given an  array containing the  [0, n] middle  nnumber nums , find  [0, n] the number in  this range that does not appear in the array.

 

Advanced:

• Can you achieve linear time complexity and only use an extra constant space algorithm to solve this problem?

 

Example 1:

Input: nums = [3,0,1]
 Output: 2
 Explanation: n = 3, because there are 3 numbers, all numbers are in the range [0,3]. 2 is the missing number because it does not appear in nums.

Example 2:

Input: nums = [0,1]
 Output: 2
 Explanation: n = 2, because there are 2 numbers, all numbers are in the range [0,2]. 2 is the missing number because it does not appear in nums.

Example 3:

Input: nums = [9,6,4,2,3,5,7,0,1]
 Output: 8
 Explanation: n = 9, because there are 9 numbers, all numbers are in the range [0,9] Inside. 8 is the missing number because it does not appear in nums.

Example 4:

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

 

prompt:

• n == nums.length
• 1 <= n <= 104
• 0 <= nums[i] <= n
• nums All numbers in are  unique

1. Sort

For the sequence after sorting, there are two special positions, the beginning and the end, which are specially considered first. Then use a for loop for the rest of the positions, the latter position should be the previous position

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
    
        if(nums[nums.size()-1] != nums.size()){
            return nums.size();
        }else if(nums[0] != 0){
            return 0;
        }
        
        for(int i = 1; i < nums.size(); i++){
            if(nums[i] != nums[i-1] + 1){
                return nums[i-1]+1;
            }
        }
        return -1;
    }
};

 

2. XOR operation

XOR operation in the official solution:

Perform an XOR operation on each number in the array for 0-n numbers. The unmissed number appears once in [0-n] and the array, and the XOR operation results in 0.

The missing value only appears once, so the final XOR result is the missing value.

This XOR operation can be used to find missing values ​​in other ordered sequences.

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int missing = nums.size();
        for(int i = 0; i < nums.size(); i++){
            missing ^= i ^ nums[i];
        }
        return missing;
    }
};

 

3. Mathematical summation

Traverse the array once, sum the indices 1 to n, and subtract the array nums[i-1].

This will not overflow, which is a good solution.

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int sum = 0;
        for(int i = 1; i <= nums.size(); i++){
            sum += i;
            sum -= nums[i-1];
        }
        return sum;
    }
};