什么可能导致我的代码给出错误的答案？ [关闭]

Question

我很好奇为什么我的代码提供了关于 QUESTION 的错误答案，而如果我制作自己的列表，它会给出正确的答案。 下面是我的代码 这是要测试的LIST

X = [12,4,4,2,2,3,1,1,1,1,1]
count = 0
#j is the second element
#it's like i+1
j = 1

for i in range(len(X)-1):
    if j == len(X):
        break
    if (X[i]>X[j]) == True:
        count+=1
    j += 1

print(count)

Answer 1

我尝试对您的代码进行最低限度的编辑，以使其输出与您链接的 Stack Overflow 帖子相同的答案。我的解决方案如下。此外，我已经评论了错误的来源。您的解决方案之所以有效，是因为在您提供的列表中，每个大于右侧整数的整数也严格大于右侧的所有整数；您的代码计算这些大于右侧整数的整数：

##### YOUR SOLUTION

X = #list
count = 0
j = 1
for i in range(len(X)-1):

    # This if condition evaluates to true if you are observing the last digit.
    # "Is the last digit greater than all digits to the right?"
    # This is trivial.
    if j == len(X):
        break

    # Is a number greater than the number immediately to the right?
    # (This does not consider all the numbers to the right, only the one directly next to it.)
    if (X[i]>X[j]) == True:
        count+=1

    # j will be each index [0, len(X) - 1] EXACTLY ONCE, not the correct behavior.
    j += 1

print(count)


##### MY SOLUTION

count = 0
j = 1
for i in range(len(X)-1):

    # Set j to be the index one greater than i
    j = i + 1

    # While, for an integer i, there exists more digits to the right.
    while j < len(X):

        # If any integer after i greater than or equal to i ...
        if X[i] <= X[j]:
            break

        # increment j
        j += 1

    # If we got to the last integer without finding an integer greater than i
    # then j now points to one index greater than those in list and increment count
    if j == len(X):
        count += 1

print(count)

Answer 2

回答您发布的链接中存在的问题的正确方法如下 -

X = [12,4,4,2,2,3,1,1,1,1,1]
count = 0

for i in range(len(X)-1): # This will loop till len(X) - 1
    flag = 0
    for j in range(i+1,len(X)): # For each element, check if it is greater than all elements on right 
        if X[j] >= X[i]:
            flag = 1
            break
    # If above condition inside for loop never passes, we can increase the count :)
    if flag == 0:
        count+=1

print(count)

输出：

3

上述解决方案将检查当前数字之后的每个数字。在您的解决方案中，您的逻辑不正确。您只使用下一个后续数字检查每个数字。但是你必须检查右边的每个后续数字（而不仅仅是 1）。

还有一个O(N) 可以解决这个问题，它比这个更有效。如果您正在寻找更优化的解决方案，您应该参考您在 O(N) 解决方案问题中发布的链接。但是，上面的代码是以幼稚的方式做的正确方法。