Add to Chrome
✅ The verified answer to this question is available below. Our community-reviewed solutions help you understand the material better.
Vaatleme programmeerimiskeeles Python kirjutatud sortimismeetodeid.
Millised neist on stabiilsed?
def sort(a): N = len(a) for i in range(N): for j in range(N-1): if a[j+1] < a[j]: a[j], a[j+1] = a[j+1], a[j]
def sort(a):
N = len(a)
for i in range(N):
for j in range(N-1):
if a[j+1] < a[j]:
a[j], a[j+1] = a[j+1], a[j]
def sort(a): N = len(a) for i in range(N): for j in range(i, 0, -1): if a[j] < a[j-1]: a[j-1], a[j] = a[j], a[j-1] else: break
for j in range(i, 0, -1):
if a[j] < a[j-1]:
a[j-1], a[j] = a[j], a[j-1]
else:
break
def sort(a): N = len(a) for i in range(N-1, 1, -1): for j in range(i): if a[i] < a[j]: a[i], a[j] = a[j], a[i]
for i in range(N-1, 1, -1):
for j in range(i):
if a[i] < a[j]:
a[i], a[j] = a[j], a[i]
def sort(a): N = len(a) for i in range(N): jmin = i for j in range(i+1, N): if a[j] < a[jmin]: jmin = j a[i], a[jmin] = a[jmin], a[i]
jmin = i
for j in range(i+1, N):
if a[j] < a[jmin]:
jmin = j
a[i], a[jmin] = a[jmin], a[i]
Get Unlimited Answers To Exam Questions - Install Crowdly Extension Now!