LeetCode 268: Missing Number

Missing number. Given an array containing n distinct numbers taken from 0, 1, 2, ... n, find the one that is missing from the array.

for example:

    Given nums = [0, 1, 3] return 2;

// This one is also bit manipulation.
int missingNumber(vector<int>& nums) {
    int n = nums.size();
    int result = 0;
    for(int i = 0; i < n; ++i) {
        result ^= i ^ nums[i];
    }
    return result ^ n;
}
// or we can use binary search if the array is already sorted.
int findMissing(vector<int>& nums) {
   int n = nums.size();
   int left = 0;
   int right = nums.size() - 1;
   while(left <= right) {
      int m = (left + right) / 2;
      if(m != 0 && nums[m-1] + 1 != nums[m]) return nums[m] - 1;
      if(m == 0 && nums[m] != 0) return 0;
      if(m !=  n - 1 && nums[m + 1] - 1 != nums[m]) return nums[m] + 1;
      if(nums[m] == m) left = m + 1;
      else right = m - 1;
   }
   return -1;
}