Careercup - Facebook面试题 - 6026101998485504

2014-05-02 10:47

题目链接

原题：

 Given an unordered array of positive integers, create an algorithm that makes sure no group of integers of size bigger than M have the same integers. 

 Input: ,,,,,,,, M =
 Output: ,,,,,,,,

题目：给定一个未排序的长度为N的整数数组，和一个正整数M。请设计算法，将N个数分成M个数一组，并且每一组都不包含重复元素。

解法：本题的要求是每个组不能出现重复的数，那么换句话说就是把重复的元素分到不同的组去。我们按打扑克时发牌的方式，将重复的牌聚在一起，然后轮流发给每个组，就能保证他们不出现在一组里了。要要重复的牌聚在一次，既可以通过排序，也可以通过哈希表来统计个数。

代码：

 // http://www.careercup.com/question?id=6026101998485504
 // Actually, I don't quite understand the problem. The poster of this problem gave it too vague..
 #include <algorithm>
 #include <iostream>
 #include <unordered_map>
 #include <vector>
 #include <xiosbase>
 using namespace std;

 class Solution {
 public:
     void disperse(vector<int> &v, int m) {
         int n = (int)v.size();

         if (n <= ) {
             return;
         }

         int count;
         unordered_map<int, int> um;

         int i;

         for (i = ; i < n; ++i) {
             ++um[v[i]];
         }

         unordered_map<int, int>::iterator umit, umit2;
         i = ;
         while (!um.empty()) {
             count = ;
             umit = um.begin();
             while (umit != um.end()) {
                 v[i++] = umit->first;
                 --(umit->second);
                 ++count;
                 if (umit->second == ) {
                     // remove empty items to speed up the search.
                     umit2 = umit;
                     ++umit;
                     um.erase(umit2);
                 } else {
                     ++umit;
                 }

                 if (count == m) {
                     // pick at most m distinct elements at each round.
                     break;
                 }
             }
         }
     };
 };

 int main()
 {
     int n, m;
     vector<int> v;
     Solution sol;
     int i;

     ios::sync_with_stdio(false);

     while (cin >> n && n > ) {
         v.resize(n);

         for (i = ; i < n; ++i) {
             cin >> v[i];
         }
         cin >> m;
         sol.disperse(v, m);
         for (i = ; i < n; ++i) {
             i ? cout << ' ',  : ;
             cout << v[i];
         }
         cout << endl;

         v.clear();
     }

     return ;
 }