
- 堆结构就是用数组实现的完全二叉树结构
- 完全二叉树中如果每棵子树的最大值都在顶部就是大根堆
- 完全二叉树中如果每棵子树的最小值都在顶部就是小根堆
- 堆结构的
heapInsert(上移) 与heapify(下沉) *** 作 - 堆结构的增大和减少
- 优先级队列结构,就是堆结构
完全二叉树:对于 i i i 节点,左孩子为 2 i + 1 2i + 1 2i+1, 右孩子为 2 i + 2 2i + 2 2i+2,父节点为 ( i − 1 ) / 2 (i - 1) / 2 (i−1)/2
2、大根堆的实现/*************************************************************************
> File Name: 012.堆.cpp
> Author: Maureen
> Mail: Maureen@qq.com
> Created Time: 三 5/11 21:09:20 2022
************************************************************************/
#include
using namespace std;
//大根堆
class MyMaxHeap {
private :
int *heap;
int limit;
int heapSize;
public:
MyMaxHeap(int limit) {
heap = new int[limit];
this->limit = limit;
heapSize = 0;
}
bool empty() {
return heapSize == 0;
}
bool isFull() {
return heapSize == limit;
}
//新加入的数,停在了index位置,依次往上移动
//移动到0位置或者比父节点值小,停止
void heapInsert(int *arr, int index) {
//包含了两个终止条件:
//index == 0 或 干不掉父节点
while (arr[index] > arr[(index - 1) / 2]) {
swap(arr, index, (index - 1) / 2);
index = (index - 1) / 2;
}
}
//从index位置往下看,不断下沉
//当较大的孩子不必index位置的数大 或者 已经没有孩子了,停止
void heapify(int *arr, int index, int heapSize) {
int left = index * 2 + 1;
while (left < heapSize) { //如果存在左孩子
//先考虑是否存在右孩子,若存在,把较大孩子的下标给largest
int largest = left + 1 < heapSize && arr[left + 1] > arr[left] ? left + 1 : left;
//孩子节点和index这个父节点比较值的大小
largest = arr[largest] > arr[index] ? largest : index;
//父节点胜出,下沉停止
if (largest == index) break;
//index和较大的孩子互换
swap(arr, largest, index);
index = largest;
left = index * 2 + 1;
}
}
void push(int value) {
if (heapSize == limit) {
cout << "The heap is full!" << endl;
return ;
}
heap[heapSize] = value;
heapInsert(heap, heapSize++);
}
//返回最大值,并且在大根堆中把最大值删掉
//剩下的数,依然保持大根堆性质
int pop() {
int ans = heap[0];
swap(heap, 0, --heapSize);
heapify(heap, 0, heapSize);
return ans;
}
void swap(int *arr, int i, int j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
void print() {
for (int i = 0; i < heapSize; i++) {
cout << heap[i] << " ";
}
cout << endl;
}
};
//for test
class RightMaxHeap {
private:
int *arr;
int limit;
int size;
public:
RightMaxHeap(int limit) {
arr = new int[limit];
this->limit = limit;
size = 0;
}
bool empty() {
return size == 0;
}
bool isFull() {
return size == limit;
}
void push(int value) {
if (size == limit) {
cout << "The heap is full" << endl;
return ;
}
arr[size++] = value;
}
int pop() {
int maxValInd = 0;
for (int i = 0; i < size; i++) {
if (arr[i] > arr[maxValInd]) {
maxValInd = i;
}
}
int ans = arr[maxValInd];
arr[maxValInd] = arr[--size];
return ans;
}
void print() {
for (int i = 0; i < size; i++) {
cout << arr[i] << " ";
}
cout << endl;
}
};
int main() {
srand(time(0));
int value = 1000;
int limit = 100;
int testTime = 1000000;
bool success = true;
for (int i = 0; i < testTime + 1; i++) {
int curLimit = rand() % limit + 1;
MyMaxHeap myHeap(curLimit);
RightMaxHeap test(curLimit);
int curOpTimes = rand() % limit;
for (int j = 0; j < curOpTimes; j++) {
if (myHeap.empty() != test.empty()) {
cout << "Oops! empty() function occurs error!" << endl;
success = false;
break;
}
if (myHeap.isFull() != test.isFull()) {
cout << "Oops! isFull() function occurs error!" << endl;
success = false;
break;
}
if (myHeap.empty()) {
int curValue = rand() % value;
myHeap.push(curValue);
test.push(curValue);
} else if (myHeap.isFull()) {
if (myHeap.pop() != test.pop()) {
cout << "Oops! pop() function occurs error!" << endl;
success = false;
break;
}
} else {
if (((rand() % 101)/ (double)101) < 0.5) {
int curValue = rand() % value;
myHeap.push(curValue);
test.push(curValue);
} else {
int ans1 = myHeap.pop();
int ans2 = test.pop();
if (ans1 != ans2) {
myHeap.print();
test.print();
cout << "ans1 = " << ans1 << ", ans2 = " << ans2 << endl;
cout << "Oops! pop() function occurs error again!" << endl;
success = false;
break;
}
}
}
}
if (!success) break;
if (i && i % 10000 == 0) { cout << i << " cases passed!" << endl; }
}
cout << "finish!" << endl;
return 0;
}
其中 heapInsert 和 heapify *** 作的时间复杂度都是
O
(
l
o
g
n
)
O(logn)
O(logn)。
问:如果数组的 0 ~ i 范围上本来已经是满足大根堆结构了,但是如果其中的某个位置 x 处的值发生了变化,如何将 0 ~ i 范围上调整为大根堆呢?
答:很简单,依次执行 heapInsert(arr, x) 和 heapify(arr, x, i + 1),这两个 *** 作中只有一个会执行。
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