Question

我不确定如何将一个项目插入到我的最大堆中，然后涓涓细流，以便最大堆属性保持不变。如果heapArray已满，则抛出异常，因此无法插入项目。

我没有使用JCF类或此程序的优先级队列。我还抛出了我的deleteMax方法，该方法删除堆的最大值并恢复堆，以便最大堆属性保持不变。

public  class MaxIntHeap {
//my global var's
private int[] heapArray; // default size of 20 in constructor
private int lastOne; //equal to size of the array(heap)
private int size;
private int k;

public void heapInsert(int v) throws HeapException {
  if(lastOne == heapArray.length - 1 ){
  throw new HeapException("The value " + v + " cannot be inserted because the heap is FULL!");
  }
  else{ //This is where i am lost as to how i should insert
     lastOne++; //size of lastOne is increased.
}

这是我的removeMax方法。

public int  removeMax ()throws HeapException  { 
  if(size == 0){
     throw new HeapException("The Heap is empty, therefore the Max value of the heap cannot be       
     removed.");
  }
  else{
     int max = heapArray[0];
     heapArray[0] = heapArray[lastOne];
     lastOne--;
     k = 0;
     while(k > 0){
        int j = k / 2;
        if(heapArray[k] < heapArray[j]){
           swap(heapArray[k], heapArray[j]); //swap method...
           k = j;

        }
        else 
           break;
     }
     return max;
   }

  }

非常感谢任何帮助。谢谢！

Answer 1

你的班级设计看起来不错。在堆中：

leftChild = 2*parent +1
rightChild = 2*parent + 2
parent = (childindex-1)/2

对于maxheap，

最后插入元素。然后将其与它进行比较家长。
如果父级大于此最新插入，则为好。
Else交换父母和这个孩子

重复直到你到达根目录。

MaxHeapImpl：

public class MaxHeap {

public int[] myHeap = new int[20];
public int begin = 0;
public int current = 0;

public int getParent(int index){
    return (index - 1)/2;
}

public int getLeftChild(int index){
    return 2*index+1;
}

public int getRighChild(int index){
    return 2*index+2;
}

public void insert(int data) {

    myHeap[current] = data;

    int i = current;
    int tmp;
    int parent;
    while(i > 0){
        parent = getParent(i);

        System.out.println(" I value"+i+" parent"+parent+" data"+data);
        if(myHeap[parent] < myHeap[i]){
            tmp = myHeap[parent];
            myHeap[parent] = myHeap[i];
            myHeap[i] = tmp;
        } else{
            break;
        }

        i = parent;

    }
    current++;
}

}

识别TestClass：

    public class MaxHeapTest {

    @Test
    public void test() {
        MaxHeap myHeap = new MaxHeap();

        myHeap.insert(40);
        myHeap.insert(20);
        myHeap.insert(10);
        myHeap.insert(25);
        myHeap.insert(30);
        myHeap.insert(100);

        for(int i = 0; i < myHeap.current;i++){
            System.out.println(" "+myHeap.myHeap[i]);
        }
    }

}

Answer 2

为解决以上问题，当前值不等于0，而是堆的大小（带有值）。

from symfit import parameters, variables, sin, cos, Fit
import numpy as np
import matplotlib.pyplot as plt
import math as m
from random import random

def f(x,a,b,omega):
    output = a[0]/2
    for k in range(1,len(a)-1):
        output = output + a[k]*m.cos(k*omega*x) + b[k]*m.sin(k*omega*x)
    return output

def monteCarlo(min,max,a,b,omega):
    x = min + random()*(max-min)
    return (x,f(x,a,b,omega))

xdata=[]
ydata=[]

#for step function between -3 and 3
#a=[5.000000e-01,1.106075e-10,2.364473e-11,1.697959e-10,2.822410e-11,3.860925e-11,-1.522386e-10]
#b=[0,6.267589e-01,2.020945e-02,1.846406e-01,3.623074e-02,8.726419e-02,4.518721e-02]
#omega=8.615115e-01

#for our spectrum with n=6
a = [-1.755069e+02,-1.097051e+02,1.571135e+02,6.491389e+01,1.220463e+02,1.778165e+02,4.564577e+01]
b = [0,-5.523502e+01,1.517397e+02,1.325198e+02,-1.015198e+02,-3.255781e+01,2.495038e+01]
omega = -5.353751e-03

#for our spectrum with n=10
#a = [-1.206104e+03,1.799223e+03,-8.465307e+02,6.798590e+02,-9.287399e+02,7.388156e+02,-2.099387e+02,-9.838123e+01,\
#    9.821148e+01,-2.827002e+01,2.393892e+00]
#b = [0,-6.084419e+02,9.311030e+02,-8.877734e+02,8.810145e+02,-1.288586e+03,1.734395e+03,-1.583775e+03,9.115344e+02,\
#     -3.045882e+02,4.586827e+01]
#omega = -1.370387e-02

#a = [0.983, 0.292, 0.110, 0.006, -0.022, -0.018, 0.006]
#b = [0.000, 0.246, 0.002, -0.021, 0.017, 0.046, 0.019]
#omega = 0.045

for n in range(0,1000):
    coor = monteCarlo(375,525,a,b,omega)
    xdata.append(coor[0])
    ydata.append(coor[1])

plt.plot(xdata, ydata, 'bo')

将项目插入Max Heap

2 个答案: