There are three methods to solve this problem: bit manipulation, rearrangement of the array, and math tricks.

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Bit Manipulation

1 class Solution {2 public:3     int missingNumber(vector
     
      & nums) {4         int ans = nums.size(), i = 0;5         for (int num : nums)6             ans ^= (num ^ (i++));7         return ans;8     }9 };
     

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Rearrangement

1 class Solution { 2 public: 3     int missingNumber(vector
     
      & nums) { 4         int n = nums.size(), r = n; 5         for (int i = 0; i < n; i++) { 6             while (nums[i] != i) { 7                 if (nums[i] == n) { 8                     r = i; 9                     break;10                 }11                 swap(nums[i], nums[nums[i]]);12             }13         }14         return r;15     }16 };
     

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Math

1 class Solution {2 public:3     int missingNumber(vector
     
      & nums) {4         int n = nums.size(), ans = n * (n + 1) / 2;5         for (int num : nums) ans -= num;6         return ans;7     }8 };