《Cracking the Coding Interview》——第18章:难题——题目9

时间:2014-04-29 14:16:52   收藏:0   阅读:423

2014-04-29 04:18

题目:有一连串的数被读入,设计一个数据结构,能随时返回当前所有数的中位数。

解法:用一个大顶堆,一个小顶堆将数分成数量最接近的两份,就能轻松得到中位数了。

代码:

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  1 // 18.9 A stream of integers are passed to you, you have to tell me the median as they keep coming in.
  2 #include <climits>
  3 #include <iostream>
  4 #include <vector>
  5 #include <queue>
  6 using namespace std;
  7 
  8 template <class T>
  9 struct LessFunctor
 10 {
 11     bool operator() (const T &x, const T &y)
 12     {
 13         return x < y;
 14     }
 15 };
 16 
 17 template <class T>
 18 struct GreaterFunctor
 19 {
 20     bool operator() (const T &x, const T &y)
 21     {
 22         return x > y;
 23     }
 24 };
 25 
 26 template <class T>
 27 class MedianArray {
 28 public:
 29     MedianArray() {
 30         n_small = 0;
 31         n_great = 0;
 32     };
 33     
 34     void push(const T& val) {
 35         if (n_great == 0) {
 36             greater_heap.push(val);
 37             ++n_great;
 38             return;
 39         }
 40         
 41         if (n_great > n_small) {
 42             smaller_heap.push(val);
 43             ++n_small;
 44         } else {
 45             greater_heap.push(val);
 46             ++n_great;
 47         }
 48         
 49         if (greater_heap.top() < smaller_heap.top()) {
 50             T tmp;
 51             
 52             tmp = greater_heap.top();
 53             greater_heap.pop();
 54             greater_heap.push(smaller_heap.top());
 55             smaller_heap.pop();
 56             smaller_heap.push(tmp);
 57         }
 58     };
 59     
 60     T median() {
 61         if (n_great == 0) {
 62             return INT_MIN;
 63         } else if (n_great > n_small) {
 64             return greater_heap.top();
 65         } else {
 66             return (smaller_heap.top() + greater_heap.top()) / 2;
 67         }
 68     };
 69     
 70     ~MedianArray() {
 71         n_small = 0;
 72         n_great = 0;
 73         while (!greater_heap.empty()) {
 74             greater_heap.pop();
 75         }
 76         while (!smaller_heap.empty()) {
 77             smaller_heap.pop();
 78         }
 79     };
 80 private:
 81     int n_small;
 82 
 83     // greater elements are stored in here.
 84     priority_queue<T, vector<T>, GreaterFunctor<T> > greater_heap;
 85 
 86     int n_great;
 87 
 88     // smaller elements are stored in here.
 89     priority_queue<T, vector<T>, LessFunctor<T> > smaller_heap;
 90 };
 91 
 92 int main()
 93 {
 94     MedianArray<int> ma;
 95     int val;
 96     
 97     while (cin >> val) {
 98         ma.push(val);
 99         cout << ma.median() << endl;
100     }
101     
102     return 0;
103 }
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《Cracking the Coding Interview》——第18章:难题——题目9,码迷,mamicode.com

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