Algorithm 查找最长回文子串(非子序列)
我正在做一个Leetcode问题,即查找最长的回文子串 例如,如果输入babad,则输出可以是bab或aba 或 如果输入为cbbd,则输出为bb 我很确定我已经弄明白了,这是我的代码Algorithm 查找最长回文子串(非子序列),algorithm,python-3.x,dynamic-programming,palindrome,Algorithm,Python 3.x,Dynamic Programming,Palindrome,我正在做一个Leetcode问题,即查找最长的回文子串 例如,如果输入babad,则输出可以是bab或aba 或 如果输入为cbbd,则输出为bb 我很确定我已经弄明白了,这是我的代码 def longestPalindrome(self, s): n = len(s) # Empty matrix. table = [[False for i in range(n)] for j in range(n)] # Identity matrix. fo
def longestPalindrome(self, s):
n = len(s)
# Empty matrix.
table = [[False for i in range(n)] for j in range(n)]
# Identity matrix.
for i in range(n):
table[i][i] = True
max_len = 0
start = 0
finish = 0
for sil in range(2, n+1):
for i in range(n-sil + 1):
j = sil + i - 1
if sil == 2:
if s[i] == s[j]:
table[i][j] = True
max_len = j-i
start = i
finish = j
else:
if s[i] == s[j] and table[i+1][j-1]:
table[i][j] = True
if (j - i) > finish-start:
max_len = j - i
start = i
finish = j
return s[start:finish+1]
它适用于大多数情况,但字符串非常长的情况除外。我正在提交我的代码,它在以下情况下失败
"esbtzjaaijqkgmtaajpsdfiqtvxsgfvijpxrvxgfumsuprzlyvhclgkhccmcnquukivlpnjlfteljvykbddtrpmxzcrdqinsnlsteonhcegtkoszzonkwjevlasgjlcquzuhdmmkhfniozhuphcfkeobturbuoefhmtgcvhlsezvkpgfebbdbhiuwdcftenihseorykdguoqotqyscwymtjejpdzqepjkadtftzwebxwyuqwyeegwxhroaaymusddwnjkvsvrwwsmolmidoybsotaqufhepinkkxicvzrgbgsarmizugbvtzfxghkhthzpuetufqvigmyhmlsgfaaqmmlblxbqxpluhaawqkdluwfirfngbhdkjjyfsxglsnakskcbsyafqpwmwmoxjwlhjduayqyzmpkmrjhbqyhongfdxmuwaqgjkcpatgbrqdllbzodnrifvhcfvgbixbwywanivsdjnbrgskyifgvksadvgzzzuogzcukskjxbohofdimkmyqypyuexypwnjlrfpbtkqyngvxjcwvngmilgwbpcsseoywetatfjijsbcekaixvqreelnlmdonknmxerjjhvmqiztsgjkijjtcyetuygqgsikxctvpxrqtuhxreidhwcklkkjayvqdzqqapgdqaapefzjfngdvjsiiivnkfimqkkucltgavwlakcfyhnpgmqxgfyjziliyqhugphhjtlllgtlcsibfdktzhcfuallqlonbsgyyvvyarvaxmchtyrtkgekkmhejwvsuumhcfcyncgeqtltfmhtlsfswaqpmwpjwgvksvazhwyrzwhyjjdbphhjcmurdcgtbvpkhbkpirhysrpcrntetacyfvgjivhaxgpqhbjahruuejdmaghoaquhiafjqaionbrjbjksxaezosxqmncejjptcksnoq"
错误消息超出了时间限制
为什么会这样?我正在做一个动态规划解决方案,这应该是一个公认的答案。您并没有过早地打破内部循环,所以在所有情况下,您仍然在做O(n²)工作 假设回文的中心必须有'xx'或'x?x'。其中x是任何出现两次的字符,并且?是任何角色 对于某些路径情况,这可能不会提高最坏情况下的运行时间,但至少在您提供的示例中,它应该可以节省大量计算