268. Missing Number

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.

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

 

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.

class Solution {
    public int missingNumber(int[] nums) { // binary search
        Arrays.sort(nums);
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] > mid)
                right = mid - 1;
            else
                left = mid + 1;
        }
        return left;
    }
}

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

public int missingNumber(int[] nums) { //sum
    int len = nums.length;
    int sum = (0+len)*(len+1)/2;
    for(int i=0; i<len; i++)
        sum-=nums[i];
    return sum;
}