常用的三种排序算法

Posted 2021-02-10 fightli

#快速排序

def QuickSort(arr,left,right):
    """
    arr: the array needed to sort
    left: the start index
    right: the end index
    """
    if(left>=right):
        return ;
    base = arr[left];
    i = left;
    j = right;
    while(i<j and arr[j]>=base):
        j = j - 1;
    arr[i] = arr[j];
    while(i<j and arr[i]<=base):
        i = i + 1;
    arr[j] = arr[i];
    arr[i] = base;
    QuickSort(arr,i+1,right);
    QuickSort(arr,0,i-1);


arr = [3,1,2,5,4,6];
QuickSort(arr,0,5)
print(arr)

　

#归并排序

#将有序的start-mid ，mid-end 两个子数组组合成从小到大排列
def MemeryArray(arr,start,mid,end):
    j = mid+1;
    i = 0;
    k = 0;
    temp = [0 for i in range(end+1)];
    while(i<=mid):
        if(arr[i]>arr[j] and j<= end):
            temp[k] = arr[j];
            j = j + 1;
        else:
            temp[k] = arr[i];
            i = i + 1;
        k = k + 1;
    
    while(k<=end):
        temp[k] = arr[j];
        k = k + 1;
        j = j + 1;
    
    for i in range(end+1):
        arr[i] = temp[i];

def MergeSort(arr,start,end):
    """
    arr:the array needed to sort
    start:the start index
    end:the end index
    """
    if(len(arr)==0 or start>=end):
        return;
    mid = int((start+end)/2);
    MergeSort(arr,start,mid);
    MergeSort(arr,mid+1,end);
    MemeryArray(arr,start,mid,end);


arr = [3,1,2,5,4,6];
MergeSort(arr,0,5)
print(arr)

　

#堆排序

def HeapAdjust(arr,i,n):
    """
    big heap
    i the current point
    n the lenght of the arr
    每次调整都是当前节点与其子节点进行交换
    """
    j = 2*i + 1; #left child
    temp = arr[i];
    while(j < n):
        if(j+1 < n and arr[j + 1] > arr[j]): #right bigger then left
            j = j + 1;
        if(arr[j] <= temp):
            break;
        arr[i] = arr[j];
        i = j;
        j = 2*i + 1;
    arr[i] = temp;


def BuildHeap(arr,n):
    i = int(n/2) - 1;
    while(i >= 0):
        HeapAdjust(arr,i,n);
        i = i - 1;

def HeapSort(arr,n):
    i = n-1;
    BuildHeap(arr,n);
    print("the original heap is {0}".format(arr));
    while(i>=1):
        arr[0],arr[i] = arr[i],arr[0];
        HeapAdjust(arr,0,i);
        print("the {0} clycle is {1}".format(i,arr));
        i = i - 1;

arr = [3,1,2,5,4,6];
HeapSort(arr,6);
print(arr)

上例堆排序使用的是大根堆，若要改为小根堆只需将：
if(j+1 < n and arr[j + 1] > arr[j]): #right bigger then left ,改为 if(j+1 < n and arr[j + 1] < arr[j]): #right smaller then left;
if(arr[j] <= temp): 改为 if(arr[j] >= temp):  
即可

给出一个应用案例

从N个数种选出最大的前n个数

#使用小根堆
def SmallRootHeap(arr,i,n):
    """
    big heap
    i the current point
    n the lenght of the arr
    每次调整都是当前节点与其子节点进行交换
    """
    j = 2*i + 1; #left child
    temp = arr[i];
    while(j < n):
        if(j+1 < n and arr[j + 1] < arr[j]): #right smaller then left
            j = j + 1;
        if(arr[j] >= temp):
            break;
        arr[i] = arr[j];
        i = j;
        j = 2*i + 1;
    arr[i] = temp;

#该问题只需要找出前n大的数，故不需要排序，只需要建立堆结构即可
def BuildSmallRootHeap(arr,n):
    i = int(n/2) - 1;
    while(i >= 0):
        SmallRootHeap(arr,i,n);
        i = i - 1;

def FindTopK(arr,n):
    """find the largest n number from the arr"""
    candidate = arr[:n];
    BuildSmallRootHeap(candidate,n);
    arr = arr[n:];
    N = len(arr);
    for i in range(N):
        try:
            if(arr[i]>candidate[0]):
                candidate[0] = arr[i];
                BuildSmallRootHeap(candidate,n);
        except:
            print(i,len(arr))  
    return candidate


#从10000个数中找出最大的100个数
largest_arr = np.random.randint(0,20000,10000);
candidate = FindTopK(largest_arr,100);
largest_arr = sorted(largest_arr,reverse=True);
largest_100 = largest_arr[:100];
candidate = sorted(candidate,reverse=True);
print(candidate);
print(largest_100);