【Algorithm】堆排，C++实现

//=================== 最小堆维护 ==================
// 调整根节点为arr[i]的子树顺序，将最小的节点放到根节点位置
// 递归调整被置换的子节点
void minHeapfy(vector<int> &arr, int i, int heapSize){
    int left = 2 * i + 1;  //对于arr[i]，它的左孩子是arr[2i+1]右孩子是arr[2i+2]
    int right = 2 * i + 2;
    int least = i;
    
    if (left < heapSize && arr[left] < arr[least])
        least = left;

    if (right < heapSize && arr[right] < arr[least]){
        least = right;
    }

    if (least != i){
        int temp = arr[i];
        arr[i] = arr[least];
        arr[least] = temp;
        minHeapfy(arr, least, heapSize);  //如果子树发生变动，递归调整有变动的子节点
    }
}

//=============== 建立最小堆 ================
// 利用minHeapfy调整每一棵子树，arr中arr[arrSize/2:]的元素都是叶子节点
void buildMinHeap(vector<int> &arr){
    int heapSize = arr.size();
    for (int i = heapSize/2-1; i >= 0; i--){
        minHeapfy(arr, i, heapSize);    
    }
}

// ================ 输出堆排结果 ====================
//由于根节点总是全堆最小的，每次置换根节点与最后一个节点
//对被置换之前的数组进行最小堆建立
void heapSort(vector<int> &arr){
    int heapSize = arr.size();
    buildMinHeap(arr);
    for (int i = heapSize - 1; i >= 0; i--){
        int temp = arr[0]; //置换根节点元素和当前数组最后一个元素
        arr[0] = arr[i];
        arr[i] = temp;

        heapSize -= 1;
        minHeapfy(arr, 0, heapSize); //对有序元素之前的节点建立最小堆
    }
}


int _tmain(int argc, _TCHAR* argv[])
{
    vector<int> primes = { 4, 1, 3, 2, 16, 9, 10, 14, 8, 7 };

    heapSort(primes);
    for (int i = 0; i < primes.size(); i++){
        cout << primes[i] << endl;
    }

    system("pause");
    return 0;
}