堆排序

堆是一种数据结构,最大堆性质:堆中的节点值总是不大于其父节点的值,堆是一颗完全二叉树。

  1 template<typename Item>
  2 class MaxHeap {
  3 private:
  4     Item *data;
  5     int count;
  6     int capacity;
  7     void shiftUp(int k) {//将新加入的元素与父节点依次比较,使之满足最大堆
  8         while (k>1 && data[k / 2] < data[k]) {
  9             swap(data[k / 2], data[k]);
 10             k /=2;
 11         }
 12     }
 13     void shiftDown(int k) {
 14         while (2 * k <= count) {
 15             int j = 2 * k; //在此轮循环中, data[k]和data[j]交换位置
 16             if (j + 1 <= count&&data[j] < data[j + 1])
 17                 j++;
 18             // data[j] 是 data[2*k]和data[2*k+1]中的最大值
 19             if (data[k] >= data[j])
 20                 break;
 21             swap(data[k], data[j]);
 22             k = j;//下一层
 23         }
 24 }
 26 public:
 27     MaxHeap(int capacity) {
 28         data = new Item[capacity + 1];//开辟的空间容量
 29         count = 0;//表示的是堆中的元素数目
 30         this->capacity = capacity;
 31     }
 32     MaxHeap(Item arr[], int n) {
 33         data = new Item[n + 1];
 34         capacity = n;
 35 
 36         for (int i = 0; i < n; i++)
 37             data[i + 1] = arr[i];
 38         count = n;
 39 
 40         for (int i = count / 2; i >= 1; i--)
 41             shiftDown(i);
 42     }
 43     ~MaxHeap() {
 44         delete[] data;
 45     }
 46     int size() {
 47         return count;
 48     }
 49     bool isEmpty() {
 50         return count == 0;
 51     }
 52     void insert(Item item) {
 53         assert(count + 1 <= capacity);
 54         data[count + 1] = item;
 55         shiftUp(count + 1);
 56         count++;
 57     }
 58     Item extractMax() {
 59         assert(count > 0);
 60         Item ret = data[1];
 61         swap(data[count], data[1]);
 62         count--;
 63         shiftDown(1);
 64 
 65         return ret;
 66     }
 67     Item getMax() {
 68         assert(count > 0);
 69         return data[1];
 70     } 71 };

堆排序:利用堆将数组进行排序,堆中的根节点存储的是最大值,由此将队中的值先插入操作,再进行去除最大值放到排序数组中,heapify过程。

 1 template<typename T>
 2 void heapSort2(T arr[], int n) {
 3 
 4     MaxHeap<T> maxheap = MaxHeap<T>(arr, n);
 5     for (int i = n - 1; i >= 0; i--)
 6         arr[i] = maxheap.extractMax();
 7 
 8 }
 9 
10 
11 template<typename T>
12 void heapSort1(T arr[], int n) {
13 
14     MaxHeap<T> maxheap = MaxHeap<T>(n);//构造一个最大堆
15     for (int i = 0; i < n; i++)
16         maxheap.insert(arr[i]);//将数组中的元素插入堆中
17 
18     for (int i = n - 1; i >= 0; i--)
19         arr[i] = maxheap.extractMax();//将堆中元素倒序插入到数组中
20 
21 }
22 //原地堆排序
23 using namespace std;
24 template<typename T>
25 void __shiftDown(T arr[], int n, int k) {
26     while (2 * k + 1 < n) {
27         int j = 2 * k + 1;
28         if (j + 1 < n&&arr[j + 1] > arr[j]) {
29             j = j + 1;
30         }
31         if (arr[k] >= arr[j]) break;
32         swap(arr[k], arr[j]);
33         k = j;
34     }
35 }
36 template <typename T>
37 void heapSort(T arr[], int n) {
38     for (int i = (n - 1) / 2; i >= 0; i--)//进行heapify操作将数组转换成堆的样子
39         __shiftDown(arr, n, i);//找到每个非叶子节点进行shiftDown;
40     for (int i = n - 1; i > 0; i--) {
41         swap(arr[0], arr[i]);//将最大的元素换到数组的最后未排序部分
42         __shiftDown(arr, i, 0);//进行shiftDown维护堆
43     }
44 }

最大索引堆:堆中存储的元素是数组的索引

  1 template<typename Item>
  2 class IndexMaxHeap {
  3 private:
  4     Item *data;
  5     int *indexes;
  6     int *reverse;
  7     int count;
  8     int capacity;
  9 
 10 
 11     void shiftUp(int k) {//交换索引
 12         while (k>1 && data[indexes[k / 2]] < data[indexes[k]]) {
 13             swap(indexes[k / 2], indexes[k]);
 14             reverse[indexes[k / 2]] = k / 2;
 15             reverse[indexes[k]] = k;
 16             k /= 2;
 17         }
 18     }
 19     void shiftDown(int k) {
 20         while (2 * k <= count) {
 21             int j = 2 * k; //在此轮循环中, data[k]和data[j]交换位置
 22             if (j + 1 <= count&&data[indexes[j]] < data[indexes[j] + 1])
 23                 j++;
 24             // data[j] 是 data[2*k]和data[2*k+1]中的最大值
 25             if (data[indexes[k]] >= data[indexes[j]])
 26                 break;
 27             swap(indexes[k], indexes[j]);
 28             reverse[indexes[k]] = k ;
 29             reverse[indexes[j]] = j;
 30             k = j;//下一层
 31         }
 32 
 33     }
 34 public:
 35     IndexMaxHeap(int capacity) {
 36         data = new Item[capacity + 1];//开辟的空间容量
 37         indexes = new int[capacity + 1];//为索引开辟空间
 38         reverse = new int[capacity + 1];
 39         for (int i = 0; i <= capacity; i++)
 40             reverse[i] = 0;
 41         count = 0;//表示的是堆中的元素数目
 42         this->capacity = capacity;
 43     }
 44     
 45     ~IndexMaxHeap() {
 46         delete[] data;
 47         delete[] indexes;
 48         delete[] reverse;
 49     }
 50     int size() {
 51         return count;
 52     }
 53     bool isEmpty() {
 54         return count == 0;
 55     }
 56     void insert(int i,Item item) {
 57         assert(i + 1 >= 1 && i + 1 < capacity);
 58         assert(count + 1 <= capacity);
 59         i += 1;
 60         indexes[count + 1] = i;
 61         reverse[count + 1] = i;
 62         data[i] = item;
 63         count++;
 64         shiftUp( count );
 65         
 66     }
 67     Item extractMax() {
 68         assert(count > 0);
 69         Item ret = data[indexes[1]];
 70         swap(indexes[1], indexes[count]);
 71         reverse[indexes[count]] = 0;
 72         reverse[indexes[1]] = 1;
 73         count--;
 74         shiftDown(1);
 75 
 76         return ret;
 77     }
 78     // 传入的i对用户而言,是从0索引的
 79     int  extractMaxIndex() {
 80         assert(count > 0);
 81         int ret = indexes[1]-1;
 82         swap(indexes[1], indexes[count]);
 83         reverse[indexes[count]] = 0;
 84         reverse[indexes[1]] = 1;
 85         count--;
 86         shiftDown(1);
 87 
 88         return ret;
 89     }
 90     Item getMax(int i) {
 91         assert(count > 0);
 92         return data[indexes[1]];
 93     }
 94     int getMaxIndex() {
 95         assert(count > 0);
 96         return indexes[1] - 1;
 97     }
 98     bool contain(int i) {
 99         assert(i + 1 >= 1 && i + 1 <= capacity);
100         return reverse[i + 1] != 0;
101     }
102 
103     Item getItem(int i) {
104         assert(contain(i));
105         return data[i + 1];
106     }
107 
108     void change(int i, Item newItem) {
109 
110         assert(contain(i));
111         i += 1;
112         data[i] = newItem;
113 
114         // 找到indexes[j] = i, j表示data[i]在堆中的位置
115         // 之后shiftUp(j), 再shiftDown(j)
116 
117         //        for( int j = 1 ; j <= count ; j ++ )
118         //            if( indexes[j] == i ){
119         //                shiftUp(j);
120         //                shiftDown(j);
121         //                return;
122         //            }
123 
124         int j = reverse[i];
125         shiftUp(j);
126         shiftDown(j);
127     }
128     // test reverse index
129     bool testReverseIndex() {
130 
131         int *copyIndexes = new int[count + 1];
132         int *copyReverseIndexes = new int[count + 1];
133 
134         for (int i = 0; i <= count; i++) {
135             copyIndexes[i] = indexes[i];
136             copyReverseIndexes[i] = reverse[i];
137         }
138 
139         copyIndexes[0] = copyReverseIndexes[0] = 0;
140         std::sort(copyIndexes, copyIndexes + count + 1);
141         std::sort(copyReverseIndexes, copyReverseIndexes + count + 1);
142 
143         bool res = true;
144         for (int i = 1; i <= count; i++)
145             if (copyIndexes[i - 1] + 1 != copyIndexes[i] || copyReverseIndexes[i - 1] + 1 != copyReverseIndexes[i])
146                 res = res || false;
147 
148         delete[] copyIndexes;
149         delete[] copyReverseIndexes;
150 
151         if (!res) {
152             cout << "Error 1" << endl;
153             return res;
154         }
155 
156         for (int i = 1; i <= count; i++)
157             if (reverse[indexes[i]] != i) {
158                 cout << "Error 2" << endl;
159                 return false;
160             }
161 
162         return true;
163     }
164 };
原文地址:https://www.cnblogs.com/bingzzzZZZ/p/8453205.html