我正在編寫一個(gè)簡單的代碼來檢查數(shù)字是否為回文。每當(dāng)數(shù)字有兩個(gè)連續(xù)的零時(shí)，print('removing',palind[-1])將列表中的錯(cuò)誤零作為目標(biāo)。n = 200314413002x = npalind = []while n > 0: d = n % 10 n = n // 10 palind.append(d) print(palind, 'is to check') actual_palind = []for i in palind: if palind[0] == palind[-1] and len(palind) % 2 == 0: print('removing',palind[0]) palind.remove(palind[0]) print('removing',palind[-1]) palind.remove(palind[-1]) print(palind,'is still a palindrome') actual_palind.append(x) else: print(x,'is not a palindrome') break print(x, 'is a palindrome')這是輸出[2, 0, 0, 3, 1, 4, 4, 1, 3, 0, 0, 2] is to checkremoving 2removing 2[0, 0, 3, 1, 4, 4, 1, 3, 0, 0] is still a palindromeremoving 0removing 0[3, 1, 4, 4, 1, 3, 0, 0] is still a palindrome200314413002 is not a palindrome200314413002 is a palindrome我錯(cuò)過了什么？
2 回答
一只甜甜圈
TA貢獻(xiàn)1836條經(jīng)驗(yàn) 獲得超5個(gè)贊
這不會(huì)從列表中刪除最后一項(xiàng)：
palind.remove(palind[-1])
它刪除列表的第一項(xiàng) equals palind[-1]，如果有多個(gè)相等的項(xiàng)目，這將是一個(gè)問題。
要?jiǎng)h除列表的最后一項(xiàng)，請執(zhí)行以下操作：
del palind[-1]
拉莫斯之舞
TA貢獻(xiàn)1820條經(jīng)驗(yàn) 獲得超10個(gè)贊
就像 zvone 在他們的回答中所說的那樣，它remove()不會(huì)根據(jù)索引刪除，而是根據(jù)值刪除。要根據(jù)索引刪除，請使用.pop()：
for i in palind:
if palind[0] == palind[-1] and len(palind) % 2 == 0:
print('removing',palind.pop(0))
print('removing',palind.pop(-1))
print(palind,'is still a palindrome')
actual_palind.append(x)
else:
print(x,'is not a palindrome')
break
輸出：
[2, 0, 0, 3, 1, 4, 4, 1, 3, 0, 0, 2] is to check
removing 2
removing 2
[0, 0, 3, 1, 4, 4, 1, 3, 0, 0] is still a palindrome
removing 0
removing 0
[0, 3, 1, 4, 4, 1, 3, 0] is still a palindrome
removing 0
removing 0
[3, 1, 4, 4, 1, 3] is still a palindrome
removing 3
removing 3
[1, 4, 4, 1] is still a palindrome
200314413002 is a palindrome
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