[Lintcode] Find the Missing number looking for lost numbers

Given An array contains N numbers of 0. N, find which number doesn ' t exist in the array.

Example

Given N = 3 [0, 1, 3] and the array, return 2 .

Challenge

Do it in-place with O (1) extra memory and O (n) time.

This question is the original question on Leetcode, please see my previous blog missing number missing. The problem with two ways to solve problems, but lintcode OJ more stringent, there is a large data set, summation will exceed the range of int, so for the solution of a word need to use long to calculate the sum of the array, the rest of the same, remember to turn the result into an int, see the code is as follows:

Solution One:

classSolution { Public:    /** * @param nums:a vector of integers * @return: an integer*/    intFindmissing (vector<int> &nums) {        //Write your code here        Longsum =0, n =nums.size ();  for(Auto &a:nums) {Sum+=A; }        return(int) (n * (n +1) *0.5-sum); }};

There is no difference between bit manipulation and previous bits, see the code below:

Solution Two:

classSolution { Public:    /** * @param nums:a vector of integers * @return: an integer*/    intFindmissing (vector<int> &nums) {        //Write your code here        intres =0;        Sort (Nums.begin (), Nums.end ());  for(inti =0; I < nums.size (); ++i) {res^= Nums[i] ^ (i +1); }        returnRes; }};

[Lintcode] Find the Missing number looking for lost numbers