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;}