所以我需要找到可以鏡像的最長子列表，知道元素 ex 的數量：n = 5my_list = [1,2,3,2,1]這是我的代碼：n = int(input())my_list = list(map(int, input().split()))c = 0s1 = my_listx = 0i = 0while i < n: s2 = s1[i:] if s2 == s2[::-1]: if c <= len(s2): c = len(s2) if i >= n-1: i = 0 n = n - 1 s1 = s1[:-1] i += 1print(c)正如我們所看到的，鏡像時列表是相同的，但是當結果不是預期的n = 10時。my_list = [1,2,3,2,1,332,6597,6416,614,31]35
2 回答
小唯快跑啊
TA貢獻1863條經驗 獲得超2個贊
我的解決方案是將每次迭代中的數組拆分為左數組和右數組，然后反轉左數組。接下來，比較每個數組中的每個元素，并在元素相同時將長度變量加一。
def longest_subarr(a):
longest_exclude = 0
for i in range(1, len(a) - 1):
# this excludes a[i] as the root
left = a[:i][::-1]
# this also excludes a[i], needs to consider this in calculation later
right = a[i + 1:]
max_length = min(len(left), len(right))
length = 0
while(length < max_length and left[length] == right[length]):
length += 1
longest_exclude = max(longest_exclude, length)
# times 2 because the current longest is for the half of the array
# plus 1 to include to root
longest_exclude = longest_exclude * 2 + 1
longest_include = 0
for i in range(1, len(a)):
# this excludes a[i] as the root
left = a[:i][::-1]
# this includes a[i]
right = a[i:]
max_length = min(len(left), len(right))
length = 0
while(length < max_length and left[length] == right[length]):
length += 1
longest_include = max(longest_include, length)
# times 2 because the current longest is for the half of the array
longest_include *= 2
return max(longest_exclude, longest_include)
print(longest_subarr([1, 4, 3, 5, 3, 4, 1]))
print(longest_subarr([1, 4, 3, 5, 5, 3, 4, 1]))
print(longest_subarr([1, 3, 2, 2, 1]))
這涵蓋了奇數長度子數組[a, b, a]和偶數長度子數組的測試用例[a, b, b, a]。
瀟湘沐
TA貢獻1816條經驗 獲得超6個贊
由于您需要可以鏡像的最長序列，因此這里有一個簡單的 O(n^2) 方法。
轉到每個索引，以它為中心，如果數字相等，則向左和向右擴展，一次一步。否則中斷，并移動到下一個索引。
def longest_mirror(my_array):
maxLength = 1
start = 0
length = len(my_array)
low = 0
high = 0
# One by one consider every character as center point of mirrored subarray
for i in range(1, length):
# checking for even length subarrays
low = i - 1
high = i
while low >= 0 and high < length and my_array[low] == my_array[high]:
if high - low + 1 > maxLength:
start = low
maxLength = high - low + 1
low -= 1
high += 1
# checking for even length subarrays
low = i - 1
high = i + 1
while low >= 0 and high < length and my_array[low] == my_array[high]:
if high - low + 1 > maxLength:
start = low
maxLength = high - low + 1
low -= 1
high += 1
return maxLength
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