如何 return python 中选择排序递归交换次数的值
How to return value of number of swaps in Selection Sort recursive in python
代码:
def selection(list, i, j, flag, swapNumber):
size = len(list)
if (i < size - 1):
if (flag):
j = i + 1;
print(list)
if (j < size):
if (list[i] > list[j]):
swapNumber +=1
print(swapNumber)
list[i],list[j]=list[j],list[i]
print(list)
selection(list, i, j + 1, 0, swapNumber);
else:
selection(list, i + 1, i+2, 1, swapNumber);
return(swapNumber)
swapNumber = 0
list = [6, 2, 3, 7, 1]
numSwap = selection(list, 0, 1, 1, swapNumber)
print(list)
print(numSwap)
所以我正在尝试 return swapNumber 的值,但是当它打印时它只显示 1。我尝试将 return(swapNumber) 放在其他地方似乎没有帮助.
有人可以帮忙吗?
传递给函数 selection
的 swapNumber
是 by value
而不是 by reference
。
一种可能的解决方案是将其声明为 global
:
def selection(list, i, j, flag):
global swapNumber
size = len(list)
if (i < size - 1):
if (flag):
j = i + 1;
print(list)
if (j < size):
if (list[i] > list[j]):
swapNumber +=1
print(swapNumber)
list[i],list[j]=list[j],list[i]
print(list)
swapNumber = selection(list, i, j + 1, 0);
else:
swapNumber = selection(list, i + 1, i+2, 1);
return(swapNumber)
swapNumber = 0
list = [6, 2, 3, 7, 1]
numSwap = selection(list, 0, 1, 1)
print(list)
print(numSwap)
一种厚颜无耻的方法是在函数内部定义一个函数并使用 nonlocal
。看起来像这样:
def selection_sort(list_):
swaps = 0
def selection(list_, i, j, flag):
nonlocal swaps
size = len(list_)
if i < size - 1:
if flag:
j = i + 1
if j < size:
if list_[i] > list_[j]:
swaps += 1
list_[i], list_[j] = list_[j], list_[i]
selection(list_, i, j + 1, 0)
else:
selection(list_, i + 1, i + 2, 1)
return swaps
return selection(_list, 0, 1, 1)
好处还在于您不需要传递 0, 1, 1
作为初始参数,只需传递要排序的列表本身。
代码:
def selection(list, i, j, flag, swapNumber):
size = len(list)
if (i < size - 1):
if (flag):
j = i + 1;
print(list)
if (j < size):
if (list[i] > list[j]):
swapNumber +=1
print(swapNumber)
list[i],list[j]=list[j],list[i]
print(list)
selection(list, i, j + 1, 0, swapNumber);
else:
selection(list, i + 1, i+2, 1, swapNumber);
return(swapNumber)
swapNumber = 0
list = [6, 2, 3, 7, 1]
numSwap = selection(list, 0, 1, 1, swapNumber)
print(list)
print(numSwap)
所以我正在尝试 return swapNumber 的值,但是当它打印时它只显示 1。我尝试将 return(swapNumber) 放在其他地方似乎没有帮助. 有人可以帮忙吗?
传递给函数 selection
的 swapNumber
是 by value
而不是 by reference
。
一种可能的解决方案是将其声明为 global
:
def selection(list, i, j, flag):
global swapNumber
size = len(list)
if (i < size - 1):
if (flag):
j = i + 1;
print(list)
if (j < size):
if (list[i] > list[j]):
swapNumber +=1
print(swapNumber)
list[i],list[j]=list[j],list[i]
print(list)
swapNumber = selection(list, i, j + 1, 0);
else:
swapNumber = selection(list, i + 1, i+2, 1);
return(swapNumber)
swapNumber = 0
list = [6, 2, 3, 7, 1]
numSwap = selection(list, 0, 1, 1)
print(list)
print(numSwap)
一种厚颜无耻的方法是在函数内部定义一个函数并使用 nonlocal
。看起来像这样:
def selection_sort(list_):
swaps = 0
def selection(list_, i, j, flag):
nonlocal swaps
size = len(list_)
if i < size - 1:
if flag:
j = i + 1
if j < size:
if list_[i] > list_[j]:
swaps += 1
list_[i], list_[j] = list_[j], list_[i]
selection(list_, i, j + 1, 0)
else:
selection(list_, i + 1, i + 2, 1)
return swaps
return selection(_list, 0, 1, 1)
好处还在于您不需要传递 0, 1, 1
作为初始参数,只需传递要排序的列表本身。