我在Codility接受了一个演示任务，我在试图弄清楚我做错了什么时遇到了一些麻烦。任务：
在Python 2.7环境中工作
给出了由N个整数组成的零索引数组A.该阵列的平衡指数是任何整数P，使得0≤P<1。 N和较低指数的元素之和等于较高指数的元素之和，即
A [0] + A [1] + ... + A [P-1] = A [P + 1] + ... + A [N-2] + A [N-1]。< / p>
假设零元素的总和等于0.如果P = 0或P = N-1，则会发生这种情况。
例如，考虑以下由N = 8个元素组成的数组A：
A[0] = -1
A[1] = 3
A[2] = -4
A[3] = 5
A[4] = 1
A[5] = -6
A[6] = 2
A[7] = 1
P = 1是该阵列的平衡指数，因为：
A[0] = −1 = A[2] + A[3] + A[4] + A[5] + A[6] + A[7]
P = 3是该阵列的平衡指数，因为：
A[0] + A[1] + A[2] = −2 = A[4] + A[5] + A[6] + A[7]
P = 7也是一个均衡指数，因为：
A[0] + A[1] + A[2] + A[3] + A[4] + A[5] + A[6] = 0
并且没有索引大于7的元素。
P = 8不是平衡指数，因为它不满足条件0≤P<1。 Ñ
写一个函数：
def解决方案（A）
给定由N个整数组成的零索引数组A，返回其任何均衡指数。如果不存在均衡指数，该函数应返回-1。
作为回应，我写了以下内容：
def solution(A):
if len(A) == 0: #If we're working with an empty list, the method should give us an empty list message and terminate there
return "Empty list, no integers to work with"
else:
equi = []
x = 0
length = len(A)
rightSum = []
leftSum = []
while x < length:
for i in A:
rightSum = A[1:i-1]
leftSum = A[i+1:length-2]
if sum(rightSum) == sum(leftSum):
equi.append(i)
return equi
else:
return -1
x += 1
pass
solution([-1,3,-4,5,1,-6,2,1])
当我编写代码时，我在测试列表中保持-1，即使我应该得到equi [1,3,7]。
另一个问题，为什么我需要通过＆＃39;方法结尾的关键字？
我应该补充一点，我对Python编码和编码非常陌生。我们将非常感谢所有人提供的任何帮助。
答案 0 :(得分：2)
考虑函数必须返回的数据类型。这是一份均衡指数清单。
这意味着函数中的每个return语句都必须返回一个列表。绝不是一个数字，绝不是一个字符串。
对于空列表，没有可能的均衡指数; return []。
当您找到均衡指数时，您可以将其附加到equi列表中，正如您所做的那样并继续。不要return循环内的while。第一个return语句结束了函数执行。
当你的循环因为你查看了所有索引而结束时，equi列表将包含循环找到的所有均衡索引。现在，在循环之后，它的时间是return equi，而不是无用的pass。
（对于奖励积分，您可以计算一次列表的总和，并注意向右移动索引会在左侧总和中添加一个元素，并从正确的总和中减去相同的元素。这样，您赢了＆＃ 39;每次都必须sum每个子列表;算法的性能将是线性的而不是二次的。）
答案 1 :(得分：2)
你的逻辑是理智的，但是你在语言的某些语法上绊倒。
我在评论时尽力帮助您的代码：
def solution(A):
if len(A) == 0: #If we're working with an empty list, the method should give us an empty list message and terminate there
return "Empty list, no integers to work with"
else:
equi = []
x = 0
length = len(A)
rightSum = []
leftSum = []
# while x < length: (removed)
# When we do for i in A, we're already iterating over each element i of the list A.
# As such, there's no need for the while loop.
for i in A:
# You switched right and left sum; elements at the 'left' are at the beginning of the list
# I also switched the name of the lists to leftList and rightList, to be more descriptive
# (a list and a sum are different things)
# I switched the i that was in the indexes to x. i is the integer on the list we're iterating over;
# its position on the list, on the other hand, is being counted with x.
leftList = A[0:x] # Go from 0, since you want to count the first element.
# We could also ommit the first index and it would begin from the first element
rightList = A[x+1:] # If we ommit the second index, it'll go until the last element
if sum(leftList) == sum(rightList):
# I changed equi.append(i) to equi.append(x), because i is the value we're iterating over, while
# x is the counter (index) of the number being currently evaluated
equi.append(x)
# return equi (removed)
# We don't want to return here. When we call return, the function exits!
# What this would do is exit the function if the sum of the left list wasn't equal to the sum of the right.
# This isn't what we want, so we'll just remove this
# else: (removed)
# return -1 (removed)
x += 1
# No pass needed; that's another thing entirely, just a nil instruction
# Now the loop is done, we have appended to equi all the equilibrium values.
# It's time to exit the function by returning the list of equi values.
# Since we must return -1 if no equilibrium indices exist, then we have to check for that as well
if len(equi) == 0:
return -1
else:
return equi
sol = solution([-1, 3, -4, 5, 1, -6, 2, 1])
print(sol) # [1, 3, 7]
还有一些其他小的东西可以提高可读性（变量命名约定，使用enumerate），但为了简洁和简单起见，我已经包括了它的工作原理。
答案 2 :(得分：0)
您只需要做的就是：
a=[-1, 3, -4, 5, 1, -6, 2, 1]
for i in range(len(a)):
print("sum of digits at Left side of the index {} is : {}" .format(i, sum(a[0:i])))
print("sum of digits at right side of the index {} is : {}" .format(i,sum(a[i+1:len(a)])))
if sum(a[0:i])==sum(a[i+1:len(a)]):
print("My Equilibirium Index is : {} ".format(i))