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: 2,1,1,1,3,4,4,4,5 M = 2 
Output: 2,1,1,3,1,4,4,5,4

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 <= 1) {
            return;
        }
        
        int count;
        unordered_map<int, int> um;
        
        int i;
        
        for (i = 0; i < n; ++i) {
            ++um[v[i]];
        }
        
        unordered_map<int, int>::iterator umit, umit2;
        i = 0;
        while (!um.empty()) {
            count = 0;
            umit = um.begin();
            while (umit != um.end()) {
                v[i++] = umit->first;
                --(umit->second);
                ++count;
                if (umit->second == 0) {
                    // 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 > 0) {
        v.resize(n);
        
        for (i = 0; i < n; ++i) {
            cin >> v[i];
        }
        cin >> m;
        sol.disperse(v, m);
        for (i = 0; i < n; ++i) {
            i ? cout << ' ', 1 : 1;
            cout << v[i];
        }
        cout << endl;
        
        v.clear();
    }
    
    return 0;
}