使用數組實現Min堆：插入和刪除Min（帶有重複項）

Question

我試圖在Java中實現Min Heap，但我遇到了插入和刪除元素的問題（在最後插入，刪除根分鐘）。它似乎大部分工作（我使用程序來直觀地顯示堆，並已經打印出新的根，當min已被刪除，類似的東西）。

我的問題是，出於某種原因，當添加一個新項目時，根不會切換爲新項目，但我無法弄清楚爲什麼。此外，似乎這只是存在大量重複時的問題，堆似乎不能完全保持順序（父項比子項小）。大多數情況下，它確實如此。只是偶爾它不會，對我來說它似乎是隨機的。

這是用泛型完成的，基本上遵循大多數算法。對於一個事實我知道的其他一切都是有效的，這對這兩種方法肯定是一個問題。

public void insert(T e) { 
    if (size == capacity) 
     increaseSize(); //this works fine 

    last = curr; //keeping track of the last index, for heapifying down/bubbling down when removing min 
    int parent = curr/2; 
    size++; //we added an element, so the size of our data set is larger 


    heap[curr] = e; //put value at end of array 

    //bubble up 
    int temp = curr; 

    while (temp > 1 && ((Comparable<T>) heap[temp]).compareTo(heap[parent]) < 0) { //if current element is less than the parent 
     //integer division 
     parent = temp/2; 
     swap(temp, parent); //the swapping method should be correct, but I included it for clarification 
     temp = parent; //just moves the index value to follow the element we added as it is bubbled up 
    } 

    curr++; //next element to be added will be after this one 


} 

public void swap(int a, int b){ 
    T temp = heap[a]; 
    heap[a] = heap[b]; 
    heap[b] = temp; 
} 


public T removeMin() { 

    //root is always min 
    T min = heap[1]; 

    //keep sure array not empty, or else size will go negative 
    if (size > 0) 
     size--; 

    //put last element as root 
    heap[1] = heap[last]; 
    heap[last] = null; 

    //keep sure array not empty, or else last will not be an index 
    if (last > 0) 
     last--; 

    //set for starting at root 
    int right = 3; 
    int left = 2; 
    int curr = 1; 
    int smaller = 0; 

    //fix heap, heapify down 
    while(left < size && right < size){ //we are in array bounds 

     if (heap[left] != null && heap[right] != null){ //so no null pointer exceptions 
      if (((Comparable<T>)heap[left]).compareTo(heap[right]) < 0) //left is smaller 
       smaller = left; 
      else if (((Comparable<T>)heap[left]).compareTo(heap[right]) > 0) //right is smaller 
       smaller = right; 
      else //they are equal 
       smaller = left; 
     } 
     if (heap[left] == null || heap[right] == null)//one child is null 
     { 
      if (heap[left] == null && heap[right] == null)//both null, stop 
       break; 
      if (heap[left] == null)//right is not null 
       smaller = right; 
      else //left is not null 
       smaller = left; 
     } 


     if (((Comparable<T>)heap[curr]).compareTo(heap[smaller]) > 0)//compare smaller or only child 
     { 
      swap(curr,smaller); //swap with child 
      curr = smaller; //so loop can check new children for new placement 
     } 
     else //if in order, stop 
      break; 

     right = 2*curr + 1; //set new children 
     left = 2*curr; 
    } 


    return min; //return root 
} 

在方法聲明的任何變量是全球性的，我知道有幾件事情可能是多餘的，就像當前整個/最後/溫度在添加的情況，所以我感到很抱歉。我試圖讓所有的名字都自我解釋，並解釋我在removeMin中做的所有檢查。任何幫助都會受到瘋狂的讚賞，我已經儘可能地尋找和調試了。我想我在這裏只是從根本上失去了一些東西。

Answer 1

只是爲了幫助您進行調試，您應該簡化代碼。 'last'變量有一些奇怪的事情發生。同樣在'插入'當你做循環，可能溫度應該爲0，也就是

while (temp >= 0 &&......