# 268. Missing Number
Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.
Example 1:
Input: [3,0,1]
Output: 2
Example 2:
Input: [9,6,4,2,3,5,7,0,1]
Output: 8
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
# Solution
Approach 1: Sorting.
Approach 2: Hash table.
Approach 3: Bit manipulation.
Approach 4: Math.
# Code (Python)
Approach 1:
Approach 2:
# Code (C++)
Approach 1:
class Solution {
public:
int missingNumber(vector<int>& nums) {
std::sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size(); ++i)
{
if (nums[i] != i)
return i;
}
return nums.size();
}
};
Approach 2:
class Solution {
public:
int missingNumber(vector<int>& nums) {
unordered_set<int> allNums;
for (int i = 0; i < nums.size(); ++i)
{
allNums.insert(nums[i]);
}
for (int i = 0; i <= nums.size(); ++i)
{
if (allNums.find(i) == allNums.end())
return i;
}
return -1;
}
};
Approach 3:
class Solution {
public:
int missingNumber(vector<int>& nums) {
int tmp = nums.size();
for (int i = 0; i < nums.size(); ++i)
{
tmp ^= nums[i] ^ i;
}
return tmp;
}
};
Approach 4:
class Solution {
public:
int missingNumber(vector<int>& nums) {
int sum = nums.size() * (nums.size() + 1) / 2; // Gauss's Formula.
for (int i = 0; i < nums.size(); ++i)
{
sum -= nums[i];
}
return sum;
}
};
class Solution {
public:
int missingNumber(vector<int>& nums) {
int sum = 0;
for (int i = 0; i < nums.size(); ++i)
{
sum += i + 1 - nums[i];
}
return sum;
}
};