《Cracking the Coding Interview》——第18章：难题——题目6

  1 // 18.6 Find the smallest one million number among one billion numbers.
  2 // Suppose one billion numbers can fit in memory.
  3 // I‘ll use quick selection algorithm to find them. This will return an unsorted result.
  4 // Time complexity is O(n), but the constant factor may be massive. I don‘t quite like this algorithm.
  5 #include <algorithm>
  6 #include <iostream>
  7 #include <vector>
  8 using namespace std;
  9 
 10 const int CUT_OFF = 3;
 11 
 12 int medianThree(vector<int> &v, int ll, int rr)
 13 {
 14     int mm = (ll + rr) / 2;
 15 
 16     if (v[ll] > v[mm]) {
 17         swap(v[ll], v[mm]);
 18     }
 19     if (v[ll] > v[rr]) {
 20         swap(v[ll], v[rr]);
 21     }
 22     if (v[mm] > v[rr]) {
 23         swap(v[mm], v[rr]);
 24     }
 25     swap(v[mm], v[rr - 1]);
 26     return v[rr - 1];
 27 }
 28 
 29 void quickSelect(vector<int> &v, int ll, int rr, int k)
 30 {
 31     // reference from "Data Structure and Algorithm Analysis in C" by Mark Allen Weiss.
 32     int pivot;
 33     int i, j;
 34     
 35     if (ll + CUT_OFF <=    rr) {
 36         pivot = medianThree(v, ll, rr);
 37         i = ll;
 38         j = rr - 1;
 39         
 40         while (true) {
 41             while (v[++i] < pivot);
 42             while (v[--j] > pivot);
 43             if (i > j) {
 44                 break;
 45             }
 46             swap(v[i], v[j]);
 47         }
 48         swap(v[i], v[rr - 1]);
 49     
 50         if (k < i) {
 51             return quickSelect(v, ll, i - 1, k);
 52         } else if (k > i) {
 53             return quickSelect(v, i + 1, rr, k);
 54         }
 55     } else {
 56         for (i = ll; i <= rr; ++i) {
 57             for (j = i + 1; j <= rr; ++j) {
 58                 if (v[i] > v[j]) {
 59                     swap(v[i], v[j]);
 60                 }
 61             }
 62         }
 63     }
 64 }
 65 
 66 int main()
 67 {
 68     vector<int> v;
 69     vector<int> res;
 70     int n, k;
 71     int i;
 72     int k_small, count;
 73     
 74     while (cin >> n >> k && (n > 0 && k > 0)) {
 75         v.resize(n);
 76         for (i = 0; i < n; ++i) {
 77             cin >> v[i];
 78         }
 79 
 80         // find the kth smallest number
 81         // this will change the order of elements
 82         quickSelect(v, 0, n - 1, k - 1);
 83         k_small = v[k - 1];
 84         count = k;
 85         for (i = 0; i < n; ++i) {
 86             if (v[i] < k_small) {
 87                 --count;
 88             }
 89         }
 90         for (i = 0; i < n; ++i) {
 91             if (v[i] < k_small) {
 92                 res.push_back(v[i]);
 93             } else if (v[i] == k_small && count > 0) {
 94                 res.push_back(v[i]);
 95                 --count;
 96             }
 97         }
 98         
 99         cout << ‘{‘;
100         for (i = 0; i < k; ++i) {
101             i ? (cout << ‘ ‘), 1 : 1;
102             cout << res[i];
103         }
104         cout << ‘}‘ << endl;
105         
106         v.clear();
107         res.clear();
108     }
109     
110     return 0;
111 }

 1 // 18.6 Find the smallest one million number among one billion numbers.
 2 // Suppose one billion numbers can fit in memory.
 3 // I‘ll use a max heap, which runs in O(n * log(k)) time, returns a sorted result.
 4 #include <iostream>
 5 #include <queue>
 6 #include <vector>
 7 using namespace std;
 8 
 9 template <class T>
10 struct myless {
11     bool operator () (const T &x, const T &y) {
12         return x < y;
13     };
14 };
15 
16 int main()
17 {
18     int val;
19     int n, k;
20     int i;
21     // max heap
22     priority_queue<int, vector<int>, myless<int> > q;
23     vector<int> v;
24     
25     while (cin >> n >> k && (n > 0 && k > 0)) {
26         k = k < n ? k : n;
27         for (i = 0; i < k; ++i) {
28             cin >> val;
29             q.push(val);
30         }
31         
32         for (i = k; i < n; ++i) {
33             cin >> val;
34             if (q.top() > val) {
35                 q.pop();
36                 q.push(val);
37             }
38         }
39         while (!q.empty()) {
40             v.push_back(q.top());
41             q.pop();
42         }
43         reverse(v.begin(), v.end());
44         
45         cout << ‘{‘;
46         for (i = 0; i < k; ++i) {
47             i ? (cout << ‘ ‘), 1 : 1;
48             cout << v[i];
49         }
50         cout << ‘}‘ << endl;
51         
52         v.clear();

53     }
54     
55     return 0;
56 }