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：
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 }
《Cracking the Coding Interview》——第18章：难题——题目9,码迷,mamicode.com
时间： 2024-12-28 00:48:12