一个堆排序问题

题目：在N个不相等的整数中找出最大的第K个数(N>K)。

思路：首先，用前K个整数构造容量为K的最小堆。然后，将后N-K个整数依次与堆顶元素比较，若比堆顶元素大，则替换堆顶元素并调整最小堆结构；反之，则继续比较下一个整数。最终，最小堆存储最大的k个数，其堆顶元素即为所求。

代码：

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <iostream>
  6 
  7 #define N 90
  8 #define K 13
  9 
 10 
 11 void HeapAdjust (int *heap, int beginidx, int endidx);
 12 void HeapConstruct (int *heap);
 13 
 14 void SetData (int *data);
 15 void ShowData (int *data);
 16 
 17 int main (int argc, char **argv)
 18 {
 19     int i;
 20     int h[N + 1];
 21 
 22     SetData(h);
 23     ShowData(h);
 24 
 25     HeapConstruct(h);
 26     for (i = K + 1; i <= N; i++)
 27     {
 28         if (h[i] > h[1])
 29         {
 30             h[1] = h[i];
 31             HeapAdjust(h, 1, K);
 32         }
 33     }
 34 
 35     printf("The Kth biggest number: %d\n", h[1]);
 36     return 0;
 37 }
 38 
 39 void HeapAdjust (int *heap, int beginidx, int endidx)
 40 {
 41     int &current = beginidx;
 42     int tmp, left, right, data = heap[current];
 43 
 44     while (left = (current << 1), left <= endidx)
 45     {
 46         right = left | 1;
 47         if ((left == endidx) || (heap[left] < heap[right]))
 48         {
 49             tmp = left;
 50         }
 51         else
 52         {
 53             tmp = right;
 54         }
 55         if (data > heap[tmp])
 56         {
 57             heap[current] = heap[tmp];
 58             current = tmp;
 59         }
 60         else
 61         {
 62             break;
 63         }
 64     }
 65     heap[current] = data;
 66 }
 67 
 68 void HeapConstruct (int *heap)
 69 {
 70     int i;
 71     for (i = K/2; i > 0; i--)
 72     {
 73         HeapAdjust(heap, i, K);
 74     }
 75 }
 76 
 77 void SetData (int *data)
 78 {
 79     bool *bdata = new bool[N + 1];
 80     memset(bdata, false, N + 1);
 81     srand(time(NULL));
 82     data[0] = -1;
 83     for (int i = 1; i <= N; i++)
 84     {
 85         data[i] = rand() % N + 1;
 86         while (bdata[data[i]])
 87         {
 88             data[i] = rand() % N + 1;
 89         }
 90         bdata[data[i]] = true;
 91     }
 92     delete []bdata;
 93 }
 94 
 95 void ShowData (int *data)
 96 {
 97     for (int i = 1; i <= N; i++)
 98     {
 99         printf("%02d ", data[i]);
100     }
101     puts("");
102 }

时间复杂度：O((N-K+1)*K*lgK).

posted @ 2016-11-22 22:43  jiu~  阅读(473)  评论(0)  编辑  收藏  举报