最坏情况为线性时间的选择算法之Python实现

算法导论第三版，9.3

import randomimport math#returns the number of elements that smaller than x#the input is A[p...r] inclusive in the convention of book,  1 <= p <= r <= n#in Python, to represent A[p...r], we should use a[p-1:r] def partition(a,p,r,x):    low = [m for m in a if m < x]    high = [m for m in a if m > x]    a[p-1:r] = low + [x] + high     return len(low)def median(a):    a.sort()    return a[(len(a)+1)/2 - 1]#x is the ith smallest element means there are (i-1) elements smaller than x#x is the 1st smallest means x is the smallestdef select(a,i):    if len(a) == 1:        return a[0]    #1.divide into n/5 groups    groups = []    numOfGroups = int(math.ceil(len(a)*1.0/5))    start,end = 0,0    for j in range(0,numOfGroups-1):        start = j*5        end = start + 5        groups.append(a[start:end])     groups.append(a[end:])    #2.find the median of each group    medians = []    for g in groups:        medians.append(median(g))    #3.find the median of the n/5 medians    x = select(medians,(len(medians)+1)/2)    #4.partition the array with x,so x is the kth smallest element     k = partition(a,1,len(a),x) + 1    #5.    if k == i :        return x     #if k is greager than i,means there are too many elements smaller than x,    #we need to narrow the range    elif k > i :        #SELECT (A[1...k],i)        return select(a[0:k],i)     #if k is not enough,we need to find the result from the other side    else:        #SELECT (A[k+1...n],i - k)        return select(a[k:],i-k)#test selectresult = []for i in range(10000):    a = random.sample(range(100), 21)    order = random.randint(1,20)    r1 = select(a,order)    a.sort()    r2 = a[order-1]    result.append(r1==r2)print(all(result))