算法培训

第一节

最长回文子串

给出一个字符串（假设长度最长为1000），求出它的最长回文子串，你可以假定只有一个满足条件的最长回文串。

样例

给出字符串 "abcdzdcab"，它的最长回文子串为 "cdzdc"。

挑战

O(n²) 时间复杂度的算法是可以接受的，如果你能用 O(n) 的算法那自然更好。

 

我的n^3代码：

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

class Solution:     """     @param s: input string     @return: the longest palindromic substring     """     def longestPalindrome(self, s):         # write your code here         start, end = 0, 0         for i in range(0, len(s)):             for j in range(i, len(s)):                 if (j-i)>(end-start) and self.isPalind(s, i, j):                     start, end = i, j         return s[start:end+1]           def isPalind(self, s, i, j):         while i<j:             if s[i] != s[j]:                 return False             i += 1             j -= 1         return True

 从中心点左右遍历的解法，注意回文是奇偶情况，这才是这个题目的关键点：

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

class Solution:     """     @param s: input string     @return: the longest palindromic substring     """     def longestPalindrome(self, s):         # write your code here         start,end = 0,0         n = len(s)           def locatePalindrome(s, i, j):             nonlocal start,end             while i>=0 and j<n:                 if s[i] == s[j]:                     if end-start < j-i+1:                         start,end = i,j+1                 else:                     break                 i -= 1                 j += 1                           for i in range(n):             # center i             locatePalindrome(s, i, i)             # center i,i+1             locatePalindrome(s, i, i+1)         return s[start:end]

 

DP解法：细节有点多，不一定能够写对。。。尤其是那个for是从大到小递减。

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

class Solution:     """     @param s: input string     @return: the longest palindromic substring     """     def longestPalindrome(self, s):         # write your code here         start = end = 0         n = len(s)         dp = [[False]*n for _ in range(n)]         for i in range(n-1, -1, -1): # for i in range(0, n) WRONG!!!!             for j in range(i, n):                 if i == j:                     dp[i][j] = True                 else:                     dp[i][j] = (s[i] == s[j]) and (j==i+1 or dp[i+1][j-1])                 if dp[i][j] and (j-i+1) > (end-start):                     start, end = i, j+1         return s[start:end]

 

Why is it not working if outside loop use for (int i = 0; i < n; i++)?

因为推导公式里 i 依赖于 i+1，所以i+1要比 i 先算出来，所以要倒过来循环。