[Leetcode 268] Missing number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n , find the one and that's missing from the array.

For example,
Given nums = [0, 1, 3] return 2 .

Note:
Your algorithm should run in linear runtime complexity. Could implement it using only constant extra space complexity?

Credits:

Special thanks to @jianchao. Li.fighter for adding the problem and creating all test cases.

First time AC code:

Class Solution{public:    int missingnumber (vector<int>& nums)    {        int sum=nums.size ();        if (sum==0)            return 0;        int temp;        for (int i=0; i<sum; ++i)        {            if (I==nums[i])                continue;            else            {                while (nums[i]!=i)                {                    if (nums[i]==sum)                    is break; else                    {                        temp=nums[nums[i]];                        Nums[nums[i]]=nums[i];                        Nums[i]=temp        ;        }}}} for (int i=0;i<sum;++i)            if (nums[i]!=i)            return i;        return sum;    }};

The second order is good ....

Class Solution{public:    int missingnumber (vector<int>& nums)    {        int sum=nums.size ();        if (sum==0)            return 0;         Sort (Nums.begin (), Nums.end ());        for (int i=0;i<sum;++i)            if (nums[i]!=i)            return i;        return sum;    }};

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

[Leetcode 268] Missing number