“1到10万这些数，去除2个并打乱次序，如何找出那两个数。（不准用位图）”

#include <algorithm>
#include <iostream>
#include <vector>
#include <ctime>
using namespace std;
void shuffle(vector<int> & v)
{
  int const N = v.size();
  for(int i = 0; i < N; ++i){
    v[i] = i + 1;
  }
  random_shuffle(v.begin(), v.end());
  cout << "target: " << v[N-2] << ' ' << v[N-1] << endl;
  v[N-2] = 0; // 乱序后，把最后两个标为删除
  v[N-1] = 0;
}
void find_missing(vector<int> & v)
{
  cout << "result: ";
  sort(v.begin(), v.end()); // 排序算不上考点，但却影响的时间复杂度。当然10万的数用O(n*Log(n))都能对付
  for(int i = 2, N = v.size(); i < N; ++i){ // v[0], v[1] is 0, so start from v[2]
    if(v[i] - v[i-1] > 1 ){ // 找不连续的
      cout << v[i] - 1 << ' ';
    }
  }
}
int main()
{
  srand(time(0));
  vector<int> v(100000);
  shuffle(v);
  find_missing(v);
}