Careercup - Facebook面试题 - 6026101998485504

2014-05-02 10:47

题目链接

原题：

1 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. 

2 

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

4 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：

 1 // http://www.careercup.com/question?id=6026101998485504

 2 // Actually, I don't quite understand the problem. The poster of this problem gave it too vague..

 3 #include <algorithm>

 4 #include <iostream>

 5 #include <unordered_map>

 6 #include <vector>

 7 #include <xiosbase>

 8 using namespace std;

 9 

10 class Solution {

11 public:

12     void disperse(vector<int> &v, int m) {

13         int n = (int)v.size();

14         

15         if (n <= 1) {

16             return;

17         }

18         

19         int count;

20         unordered_map<int, int> um;

21         

22         int i;

23         

24         for (i = 0; i < n; ++i) {

25             ++um[v[i]];

26         }

27         

28         unordered_map<int, int>::iterator umit, umit2;

29         i = 0;

30         while (!um.empty()) {

31             count = 0;

32             umit = um.begin();

33             while (umit != um.end()) {

34                 v[i++] = umit->first;

35                 --(umit->second);

36                 ++count;

37                 if (umit->second == 0) {

38                     // remove empty items to speed up the search.

39                     umit2 = umit;

40                     ++umit;

41                     um.erase(umit2);

42                 } else {

43                     ++umit;

44                 }

45                 

46                 if (count == m) {

47                     // pick at most m distinct elements at each round.

48                     break;

49                 }

50             }

51         }

52     };

53 };

54 

55 int main()

56 {

57     int n, m;

58     vector<int> v;

59     Solution sol;

60     int i;

61     

62     ios::sync_with_stdio(false);

63     

64     while (cin >> n && n > 0) {

65         v.resize(n);

66         

67         for (i = 0; i < n; ++i) {

68             cin >> v[i];

69         }

70         cin >> m;

71         sol.disperse(v, m);

72         for (i = 0; i < n; ++i) {

73             i ? cout << ' ', 1 : 1;

74             cout << v[i];

75         }

76         cout << endl;

77         

78         v.clear();

79     }

80     

81     return 0;

82 }